Class 10 RD SHARMA Solutions Maths Chapter 11 - Trigonometric Identities

Ex. 11.1

Ex. 11.2

11.56

11.57

11.58

11.59

Trigonometric Identities Exercise Ex. 11.1

Solution 1

Solution 2

Solution 3

Solution 4

Solution 5

Solution 6

Solution 7

Solution 8

Solution 9

Solution 10

Solution 11

Solution 12

Solution 13

Solution 14

Solution 15

Solution 16

Solution 17

Solution 18

Solution 19

Solution 20

Solution 21

Solution 22

Solution 23(i)

Solution 23(ii)

Solution 24

Solution 25

Solution 26

Solution 27

Solution 28

Solution 29

Solution 30

Solution 31

Solution 32

Solution 33

Solution 34

Solution 35

Solution 36

Solution 37. i

Solution 37. ii

L.H.S.,

Solution 38(i)

Solution 38(ii)

Solution 38(iii)

Solution 38(iv)

Solution 39

Solution 40

Solution 41

Solution 42

Solution 43

Solution 44

Solution 45

Solution 46

Solution 47 (i)

Solution 47 (ii)

Solution 47 (iii)

Solution 47(iv)

Solution 48

Solution 49

Solution 50

Solution 51

Solution 52

Solution 53

Solution 54

Solution 55. i

Solution 55. ii

Solution 56

Solution 57

Solution 58

Solution 59

Solution 60

Solution 61

Solution 62

Solution 63

Solution 64

Solution 65. i

Solution 65(ii)

Solution 66

Solution 67

Solution 68

Solution 69

Solution 70

Solution 71

Solution 72

Solution 73

Solution 74

Solution 75

Solution 76

Solution 77

Solution 78

Solution 79

Solution 80

Solution 81

Solution 82

Solution 83

Solution 84

Solution 85

Solution 86

Solution 23 (iii)

To prove:

Consider LHS

= RHS

Hence, proved.

Trigonometric Identities Exercise Ex. 11.2

Solution 1

Solution 2

Solution 3

Solution 5

Solution 6

Solution 7

Solution 8

Solution 9

Solution 10

Solution 11

Solution 12

Solution 13

Solution 14

Solution 4

Given:

Consider,

… [Multiplying and dividing by tan θ]

Trigonometric Identities Exercise 11.56

Solution 1

$begin mathsize 11px style We space know table attributes columnalign left end attributes row cell text tan end text to the power of text 2 end text end exponent straight theta text = sec end text to the power of text 2 end text end exponent straight theta minus 1 end cell row cell text end text rightwards double arrow text sec end text to the power of text 2 end text end exponent straight theta minus tan squared straight theta equals 1 end cell row cell text end text rightwards double arrow left parenthesis secθ minus tanθ right parenthesis text end text left parenthesis secθ plus tanθ right parenthesis equals 1 end cell row cell text end text rightwards double arrow secθ minus tanθ equals fraction numerator 1 over denominator secθ plus tanθ end fraction text -- end text box enclose 1 end cell row cell Given text sec end text straight theta text + tan end text straight theta text = x --- end text box enclose 2 end cell row cell so comma text sec end text straight theta text - tan end text straight theta text = end text 1 over straight x text --- end text box enclose 3 end cell row cell Apply text end text box enclose 2 text + end text box enclose 3 end cell row cell rightwards double arrow 2 secθ equals straight x plus 1 over straight x end cell row cell rightwards double arrow 2 secθ equals fraction numerator straight x squared plus 1 over denominator straight x end fraction end cell row cell rightwards double arrow box enclose secθ equals fraction numerator straight x squared plus 1 over denominator 2 straight x end fraction end enclose end cell end table So comma space the space correct space option space is space left parenthesis straight b right parenthesis. end style$

Solution 2

$begin mathsize 11px style table attributes columnalign left end attributes row cell Given space secθ + tanθ equals space straight x space space space space space minus negative negative box enclose 1 end cell row cell text ⇒ end text fraction numerator 1 over denominator space secθ space plus space tanθ end fraction equals space 1 over straight x end cell row cell text ⇒ end text fraction numerator space secθ space minus space tanθ over denominator space secθ squared space minus space tanθ squared end fraction equals space 1 over straight x end cell row cell text ⇒ end text space secθ space minus space tanθ equals space 1 over straight x.... left parenthesis Since space space secθ squared space equals 1 plus space tanθ squared right parenthesis end cell end table space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space minus negative negative box enclose 2 table attributes columnalign left end attributes row cell Apply straight space box enclose 1 - box enclose 2 end cell row cell rightwards double arrow space space 2 tanθ space equals space straight x minus 1 over straight x end cell row cell rightwards double arrow straight space box enclose tanθ equals fraction numerator straight x squared minus 1 over denominator 2 straight x end fraction end enclose end cell end table So comma space the space correct space option space is space left parenthesis straight d right parenthesis. end style$

Solution 3

$begin mathsize 11px style We space know table attributes columnalign left end attributes row cell text end text sin squared straight theta plus cos squared equals 1 end cell row cell rightwards double arrow text end text box enclose cos squared equals 1 minus sin squared straight theta end enclose text --- end text box enclose 1 end cell row cell Given text end text square root of fraction numerator 1 plus sinθ over denominator 1 minus sinθ end fraction end root end cell row cell text end text rightwards double arrow text end text square root of fraction numerator left parenthesis 1 plus sinθ right parenthesis text end text left parenthesis 1 plus sinθ right parenthesis over denominator left parenthesis 1 minus sinθ right parenthesis text end text left parenthesis 1 plus sinθ right parenthesis end fraction end root end cell row cell text end text rightwards double arrow text end text square root of fraction numerator left parenthesis 1 plus sinθ right parenthesis squared over denominator 1 minus sin squared straight theta end fraction end root end cell row cell from text end text box enclose 1 end cell row cell text end text rightwards double arrow text end text square root of fraction numerator left parenthesis 1 plus sinθ right parenthesis squared over denominator cos squared straight theta end fraction end root text end text end cell row cell text end text rightwards double arrow text end text fraction numerator 1 plus sinθ over denominator cosθ end fraction end cell row cell text end text rightwards double arrow text end text 1 over cosθ text end text plus text end text sinθ over cosθ end cell row cell text end text rightwards double arrow text sec end text straight theta text + tan end text straight theta end cell end table So comma space the space correct space option space is space left parenthesis straight a right parenthesis. end style$

Solution 4

$begin mathsize 11px style We space know table attributes columnalign left end attributes row cell text cos end text squared straight theta plus sin squared equals 1 end cell row cell rightwards double arrow text end text box enclose sin squared straight theta equals 1 minus cso squared straight theta end enclose text --- end text box enclose 1 end cell row cell Given text end text square root of fraction numerator 1 plus cosθ over denominator 1 minus cosθ end fraction end root end cell row cell text end text equals text end text square root of fraction numerator left parenthesis 1 plus cosθ right parenthesis text end text left parenthesis 1 plus cosθ right parenthesis over denominator left parenthesis 1 minus cosθ right parenthesis text end text left parenthesis 1 plus cosθ right parenthesis end fraction end root end cell row cell text end text equals text end text square root of fraction numerator left parenthesis 1 plus cosθ right parenthesis squared over denominator 1 minus cos squared straight theta end fraction end root end cell row cell text end text equals text end text square root of fraction numerator left parenthesis 1 plus cosθ right parenthesis squared over denominator sin squared straight theta end fraction end root text from equation end text box enclose 1 end cell row cell text end text equals text end text fraction numerator 1 plus cosθ over denominator sinθ end fraction end cell row cell text end text equals text end text 1 over sinθ text end text plus text end text cosθ over sinθ end cell row cell text end text equals text cosec end text straight theta text + cot end text straight theta end cell end table So comma space the space correct space option space is space left parenthesis straight b right parenthesis. end style$

Solution 5

$begin mathsize 11px style We space know table attributes columnalign left end attributes row cell text end text 1 text + tan end text to the power of text 2 end text end exponent straight A text = sec end text to the power of text 2 end text end exponent straight A text --- end text box enclose 1 end cell row cell rightwards double arrow text sec end text to the power of text 2 end text end exponent straight A minus 1 equals tan squared straight A text --- end text box enclose 2 end cell row cell Given colon text sec end text to the power of text 4 end text end exponent straight A minus sec squared text A end text end cell row cell text end text equals text sec end text to the power of text 2 end text end exponent straight A left parenthesis sec squared text A end text minus 1 right parenthesis end cell row cell text from end text box enclose 1 text & end text box enclose 2 end cell row cell text end text rightwards double arrow text end text left parenthesis 1 plus tan squared straight A right parenthesis tan squared straight A end cell row cell text end text equals text tan end text to the power of text 2 end text end exponent straight A plus text tan end text to the power of 4 straight A end cell end table So comma space the space correct space option space is space left parenthesis straight c right parenthesis. end style$

Solution 6

$begin mathsize 11px style We space space know table attributes columnalign left end attributes row cell text end text sin squared straight A plus cos squared straight A equals 1 text --- end text box enclose 1 end cell row cell Given text cos end text to the power of text 4 end text end exponent straight A minus sin to the power of 4 straight A end cell row cell text end text equals text end text left parenthesis cos squared straight A right parenthesis squared minus left parenthesis sin squared straight A right parenthesis squared text end text end cell row cell text end text equals text end text left parenthesis cos squared straight A plus sin squared straight A right parenthesis text end text left parenthesis cos squared straight A minus sin squared straight A right parenthesis end cell row cell from text end text box enclose 1 end cell row cell text end text rightwards double arrow text cos end text to the power of text 2 end text end exponent straight A minus sin squared straight A text end text left curly bracket table attributes columnalign left end attributes row cell straight a squared minus straight b squared end cell row cell equals left parenthesis straight a minus straight b left parenthesis straight a plus straight b right parenthesis right parenthesis end cell end table right curly bracket end cell row cell text end text equals text cos end text to the power of text 2 end text end exponent straight A minus left parenthesis text 1- cos end text to the power of text 2 end text end exponent straight A right parenthesis end cell row cell text end text equals text cos end text to the power of text 2 end text end exponent straight A minus text 1 +cos end text to the power of text 2 end text end exponent straight A end cell row cell text end text equals text 2cos end text to the power of text 2 end text end exponent straight A minus 1 text end text end cell end table So comma space the space correct space option space is space left parenthesis straight b right parenthesis. end style$

Trigonometric Identities Exercise 11.57

Solution 7

$begin mathsize 11px style We space know table attributes columnalign left end attributes row cell text sin end text to the power of text 2 end text end exponent straight theta plus cos squared straight theta equals 1 end cell row cell text end text rightwards double arrow text 1- end text cos squared straight theta equals text sin end text to the power of text 2 end text end exponent straight theta text --- end text box enclose 1 text end text end cell row cell text Given : end text fraction numerator sinθ over denominator 1 plus cosθ end fraction end cell row cell text end text equals text end text fraction numerator sinθ left parenthesis 1 minus cosθ right parenthesis over denominator left parenthesis 1 plus cosθ right parenthesis text end text left parenthesis 1 minus cosθ right parenthesis end fraction end cell row cell text end text equals text end text fraction numerator sinθ left parenthesis 1 minus cosθ right parenthesis over denominator text 1-cos end text to the power of text 2 end text end exponent straight theta end fraction end cell row cell from text end text box enclose 1 end cell row cell text end text rightwards double arrow text end text fraction numerator sinθ left parenthesis 1 minus cosθ right parenthesis over denominator text sin end text to the power of text 2 end text end exponent straight theta end fraction end cell row cell text end text equals text end text fraction numerator 1 minus cosθ over denominator sinθ end fraction end cell end table So comma space the space correct space option space is space left parenthesis straight c right parenthesis. end style$

Solution 8

$begin mathsize 11px style We space know table attributes columnalign left end attributes row cell text tan end text straight theta text = end text sinθ over cosθ text --- end text box enclose 1 end cell row cell text cot end text straight theta text = end text cosθ over sinθ text --- end text box enclose 2 end cell row cell Given colon text end text fraction numerator sinθ over denominator 1 minus cosθ end fraction plus fraction numerator cosθ over denominator 1 minus tanθ end fraction end cell row cell from text end text box enclose 1 text & end text box enclose 2 end cell row cell text end text rightwards double arrow text end text fraction numerator sinθ over denominator 1 minus cosθ over sinθ end fraction text end text plus text end text fraction numerator cosθ over denominator 1 minus sinθ over cosθ end fraction end cell row cell text end text equals text end text fraction numerator sin squared straight theta over denominator sinθ minus cosθ end fraction text end text plus text end text fraction numerator cos squared straight theta over denominator cosθ minus sinθ end fraction end cell row cell text end text equals text end text fraction numerator sin squared straight theta minus text end text cos squared straight theta text end text over denominator sinθ minus text end text cosθ text end text end fraction end cell row cell text end text equals text end text fraction numerator left parenthesis sinθ minus up diagonal strike straight c osθ right parenthesis text end text left parenthesis sinθ plus cosθ right parenthesis over denominator left parenthesis sinθ minus up diagonal strike straight c osθ right parenthesis end fraction end cell row cell text end text equals text end text left parenthesis sinθ plus text cos end text straight theta right parenthesis end cell end table So comma space the space correct space option space is space left parenthesis straight c right parenthesis. end style$

Solution 9

$begin mathsize 11px style We space know table attributes columnalign left end attributes row cell cotθ equals text end text cosθ over sinθ text --- end text box enclose 1 text end text cosecθ equals text end text 1 over sinθ text --- end text box enclose 3 end cell row cell tanθ equals text end text sinθ over cosθ text --- end text box enclose 2 text end text secθ equals text end text 1 over cosθ text --- end text box enclose 4 text end text end cell row cell text Given : end text left parenthesis 1 plus cotθ minus text cosec end text straight theta right parenthesis text end text left parenthesis 1 plus tanθ plus secθ right parenthesis text end text end cell row cell from text end text box enclose 1 text , end text box enclose 2 text , end text box enclose 3 text , end text box enclose 4 end cell row cell rightwards double arrow left parenthesis 1 plus cosθ over sinθ minus 1 over sinθ right parenthesis text end text left parenthesis 1 plus sinθ over cosθ plus 1 over cosθ right parenthesis end cell row cell rightwards double arrow left parenthesis fraction numerator sinθ plus cosθ minus 1 over denominator sinθ end fraction right parenthesis text end text left parenthesis fraction numerator sinθ plus cosθ plus 1 over denominator cosθ end fraction right parenthesis end cell row cell rightwards double arrow fraction numerator left parenthesis sinθ plus cosθ right parenthesis squared minus 1 over denominator sinθcosθ end fraction text end text left curly bracket left parenthesis straight a plus straight b right parenthesis left parenthesis straight a minus straight b right parenthesis equals straight a squared minus straight b squared right curly bracket end cell row cell rightwards double arrow fraction numerator sin squared straight theta plus cos squared plus 2 sinθcosθ minus 1 over denominator sinθcosθ end fraction equals 2 text end text left curly bracket sin squared straight theta plus cos squared straight theta equals 1 right curly bracket end cell row cell rightwards double arrow fraction numerator 1 plus 2 sinθcosθ minus 1 over denominator sinθcosθ end fraction equals 2 end cell end table So comma space the space correct space option space is space left parenthesis straight b right parenthesis. end style$

Solution 10

$begin mathsize 11px style table attributes columnalign left end attributes row cell text end text fraction numerator tanθ over denominator secθ minus 1 end fraction plus fraction numerator tanθ over denominator secθ plus 1 end fraction end cell row cell rightwards double arrow text end text fraction numerator tanθ left parenthesis secθ plus 1 right parenthesis plus tanθ left parenthesis secθ minus 1 right parenthesis over denominator left parenthesis secθ minus 1 right parenthesis left parenthesis secθ plus 1 right parenthesis end fraction text end text end cell row cell rightwards double arrow text end text fraction numerator tanθ text end text secθ plus tanθ plus tanθ text end text secθ minus tanθ over denominator sec squared straight theta minus 1 end fraction end cell row cell rightwards double arrow text end text fraction numerator 2 tanθsecθ over denominator tan squared straight theta end fraction text end text left curly bracket sec squared straight theta equals 1 plus tan squared straight theta right curly bracket end cell row cell rightwards double arrow text end text fraction numerator 2 secθ over denominator tanθ end fraction end cell row cell rightwards double arrow text end text fraction numerator 2 left parenthesis 1 over cosθ right parenthesis over denominator sinθ over cosθ end fraction end cell row cell rightwards double arrow text end text 2 over sinθ equals 2 cosecθ end cell end table So comma space the space correct space option space is space left parenthesis straight c right parenthesis. end style$

Solution 11

$begin mathsize 11px style table attributes columnalign left end attributes row cell We text know end text box enclose sin squared straight theta plus cos squared straight theta plus 1 end enclose text --- end text box enclose 1 end cell row cell text end text left parenthesis cosecθ minus sinθ right parenthesis left parenthesis secθ minus cosθ right parenthesis left parenthesis tanθ plus cotθ right parenthesis end cell row cell equals text end text left parenthesis 1 over sinθ minus sinθ right parenthesis left parenthesis 1 over cosθ minus cosθ right parenthesis left parenthesis sinθ over cosθ plus cosθ over sinθ right parenthesis end cell row cell equals text end text fraction numerator left parenthesis 1 minus sin squared straight theta right parenthesis over denominator sinθ end fraction cross times fraction numerator left parenthesis 1 minus cos squared straight theta right parenthesis over denominator cosθ end fraction cross times fraction numerator left parenthesis sinθ plus cos squared straight theta right parenthesis over denominator sinθcosθ end fraction end cell row cell from text end text box enclose 1 end cell row cell rightwards double arrow text end text fraction numerator cos squared straight theta over denominator sinθ end fraction cross times fraction numerator sin squared straight theta over denominator cosθ end fraction cross times 1 over sinθcosθ end cell row cell equals text 1 end text end cell end table So comma space the space correct space option space is space left parenthesis straight b right parenthesis. end style$

Solution 12

$begin mathsize 11px style Given space straight x equals acosθ space space space space and space space space straight y equals bsinθ table attributes columnalign left end attributes row cell space space space space straight x squared equals straight a squared cos squared straight theta space space space space and space space space straight y to the power of straight 2 equals straight b squared sin squared straight theta end cell row cell rightwards double arrow space straight b to the power of straight 2 straight x squared equals straight a squared straight b to the power of straight 2 cos squared straight theta space space space and space space straight a squared straight y to the power of straight 2 equals straight a squared straight b to the power of straight 2 sin squared straight theta end cell row cell rightwards double arrow space straight b to the power of straight 2 straight x squared plus straight space straight a squared straight y to the power of straight 2 equals straight space straight a squared straight b to the power of straight 2 straight space left parenthesis sin squared straight theta plus cos squared straight theta right parenthesis end cell row cell rightwards double arrow space straight b to the power of straight 2 straight x squared plus straight space straight a squared straight y to the power of straight 2 equals straight space straight a squared straight b to the power of straight 2 left parenthesis 1 right parenthesis end cell row cell space space space space space space space space space space space space space space space space space space space space space space space space space space equals space straight a squared straight b to the power of straight 2 end cell end table So comma space the space correct space option space is space left parenthesis straight a right parenthesis. end style$

Solution 13

$begin mathsize 11px style table attributes columnalign left end attributes row cell We straight space know space space end cell row cell sec squared straight theta straight space – straight space tan squared straight theta equals 1 space space --- box enclose 1 end cell row cell straight space rightwards double arrow space straight b to the power of straight 2 straight x squared straight space minus space straight a to the power of straight 2 straight y squared end cell row cell straight space equals space straight b to the power of straight 2 left parenthesis straight a straight space secθ right parenthesis squared minus space straight a to the power of straight 2 left parenthesis straight b straight space tanθ right parenthesis squared end cell row cell straight space equals space straight a to the power of straight 2 straight b to the power of straight 2 straight space sec squared straight theta minus straight a to the power of straight 2 straight b to the power of straight 2 straight space tan squared straight theta end cell row cell straight space equals space straight a to the power of straight 2 straight b to the power of straight 2 straight space left parenthesis sec squared straight theta straight space – straight space tan squared straight theta right parenthesis end cell row cell straight space space from space box enclose 1 end cell row cell straight space rightwards double arrow space straight a to the power of straight 2 straight b to the power of straight 2 straight space left parenthesis sec squared straight theta straight space – straight space tan squared straight theta right parenthesis equals space straight a to the power of straight 2 straight b to the power of straight 2 straight space end cell end table So comma space the space correct space option space is space left parenthesis straight d right parenthesis. end style$

Solution 14

$begin mathsize 11px style We space know table attributes columnalign left end attributes row cell text cot 3 end text straight theta text = end text fraction numerator 3 cotθ minus cot cubed straight theta over denominator 1 minus 3 cot squared straight theta end fraction text ---- end text box enclose 1 end cell row cell text tan 3 end text straight theta text = end text fraction numerator 3 tanθ minus tan cubed straight theta over denominator 1 minus 3 tan squared straight theta end fraction text ---- end text box enclose 2 end cell row cell rightwards double arrow text end text fraction numerator cotθ over denominator cotθ minus cot 3 straight theta end fraction text end text plus text end text fraction numerator tanθ over denominator tanθ minus tan 3 straight theta end fraction end cell row cell from text end text box enclose 1 text & end text box enclose 2 text end text end cell row cell rightwards double arrow text end text fraction numerator cotθ over denominator cotθ minus left parenthesis fraction numerator 3 cotθ minus cot cubed straight theta over denominator 1 minus 3 cot squared straight theta end fraction right parenthesis end fraction text end text plus text end text fraction numerator tanθ over denominator tanθ minus left parenthesis fraction numerator 3 tanθ minus tan cubed straight theta over denominator 1 minus 3 tan squared straight theta end fraction right parenthesis end fraction end cell row cell equals text end text fraction numerator cotθ left parenthesis 1 minus 3 cot squared straight theta right parenthesis over denominator negative 2 cotθ minus 2 cot cubed straight theta end fraction text + end text fraction numerator tanθ left parenthesis 1 minus 3 tan squared straight theta right parenthesis over denominator negative 2 tanθ minus 2 tan cubed straight theta end fraction end cell row cell equals text - end text fraction numerator left parenthesis 1 minus 3 tan squared straight theta right parenthesis over denominator 2 left parenthesis 1 plus tan squared straight theta right parenthesis end fraction text end text minus text end text fraction numerator left parenthesis 1 minus 3 cot squared straight theta right parenthesis over denominator 2 left parenthesis 1 plus cot squared straight theta right parenthesis end fraction end cell row cell equals text - end text 1 half left curly bracket fraction numerator 1 minus 3 tan squared straight theta over denominator sec squared straight theta end fraction text + end text fraction numerator 1 minus 3 cot squared straight theta over denominator cosec squared straight theta end fraction text end text right curly bracket end cell row cell equals text - end text 1 half left curly bracket fraction numerator 1 minus fraction numerator 3 sin squared straight theta over denominator cos squared straight theta end fraction over denominator fraction numerator 1 over denominator cos squared straight theta end fraction end fraction text + end text fraction numerator 1 minus fraction numerator 3 cos squared straight theta over denominator sin squared straight theta end fraction over denominator fraction numerator 1 over denominator cos squared straight theta end fraction end fraction text end text right curly bracket end cell row cell equals text - end text 1 half left curly bracket cosθ minus 3 sin squared straight theta text + sin end text to the power of text 2 end text end exponent straight theta minus 3 cos squared straight theta right curly bracket end cell row cell equals text - end text 1 half left curly bracket negative 2 cos squared straight theta text - 2sin end text to the power of text 2 end text end exponent straight theta right curly bracket end cell row cell equals text sin end text to the power of text 2 end text end exponent straight theta text + cos end text to the power of text 2 end text end exponent straight theta end cell row cell equals text 1 end text end cell end table So comma space the space correct space option space is space left parenthesis straight b right parenthesis. end style$

Solution 15

$Error converting from MathML to accessible text.$

Solution 16

$begin mathsize 11px style table attributes columnalign left end attributes row cell acosθ plus bsinθ equals 4 end cell row cell On text squaring both sides end text end cell row cell rightwards double arrow text a end text to the power of text 2 end text end exponent cos squared straight theta plus straight b squared sin squared straight theta plus 2 absinθcosθ equals 16 text --- end text box enclose 1 end cell row cell text asin end text straight theta text - bcos end text straight theta equals 3 end cell row cell text On squaring both sides end text end cell row cell rightwards double arrow text a end text to the power of text 2 end text end exponent sin squared straight theta plus straight b squared cos squared straight theta minus 2 absincosθ equals 9 text --- end text box enclose 2 end cell row cell Adding text end text box enclose 1 text & end text box enclose 2 text end text end cell row cell rightwards double arrow text a end text to the power of text 2 end text end exponent left parenthesis sin squared straight theta plus cos squared straight theta right parenthesis plus straight b squared left parenthesis sin squared straight theta plus cos squared straight theta right parenthesis equals 25 end cell row cell rightwards double arrow text a end text to the power of text 2 end text end exponent plus straight b squared equals 25 end cell end table So comma space the space correct space option space is space left parenthesis straight c right parenthesis. end style$

Solution 17

$begin mathsize 11px style table attributes columnalign left end attributes row cell Given straight space acotθ plus bcosecθ equals straight p end cell row cell Squaring straight space both space sides end cell row cell rightwards double arrow space straight a to the power of straight 2 cot squared straight theta plus straight b squared cosec squared straight theta plus 2 abcotθcosecθ equals straight p squared --- box enclose 1 end cell row cell Given space bcotθ + acosecθ equals straight q end cell row cell Squaring space both space sides end cell row cell rightwards double arrow space straight b to the power of straight 2 cot squared straight theta plus straight a squared cosec squared straight theta plus 2 abcotθcoseθ equals straight q squared --- box enclose 2 end cell row cell Subtracting space space box enclose 2 space from space box enclose 1 comma end cell row cell rightwards double arrow space straight a to the power of straight 2 left parenthesis cot squared straight theta minus cosec squared straight theta right parenthesis plus straight b squared left parenthesis cosec squared straight theta minus cot squared straight theta right parenthesis equals straight p squared minus straight q squared --- box enclose 3 end cell row cell We space know space space cosec squared straight theta equals 1 plus straight space cot squared straight theta end cell row cell space rightwards double arrow cosec squared straight theta minus cot squared straight theta equals 1 --- box enclose 4 end cell row cell from space box enclose 3 & box enclose 4 end cell row cell rightwards double arrow negative straight a to the power of straight 2 plus straight b to the power of straight 2 equals straight p squared minus straight q squared end cell row cell rightwards double arrow space space straight p squared minus straight q squared equals straight b to the power of straight 2 minus straight a to the power of straight 2 end cell end table So comma space the space correct space option space is space left parenthesis straight b right parenthesis. end style$

Solution 18

$begin mathsize 11px style We space know table attributes columnalign left end attributes row cell text end text sin left parenthesis 90 minus straight theta right parenthesis equals cos left parenthesis straight theta right parenthesis end cell row cell rightwards double arrow text for end text straight theta text = 29 end text to the power of text 0 end text end exponent end cell row cell rightwards double arrow text sin end text left parenthesis 61 to the power of 0 right parenthesis equals cos 29 to the power of 0 text --- end text box enclose 1 end cell row cell text end text also text sin end text to the power of text 2 end text end exponent straight theta plus cos squared straight theta equals 1 text --- end text box enclose 2 end cell row cell rightwards double arrow text sin end text to the power of text 2 end text end exponent text 29 end text to the power of text 0 end text end exponent plus sin squared 61 to the power of 0 end cell row cell text end text from text end text box enclose 1 end cell row cell rightwards double arrow text sin end text to the power of text 2 end text end exponent text 29 end text to the power of text 0 end text end exponent plus cos to the power of text 2 end text end exponent text 29 end text to the power of text 0 end text end exponent end cell row cell rightwards double arrow text 1 end text end cell end table So comma space the space correct space option space is space left parenthesis straight a right parenthesis. end style$

Solution 19

$begin mathsize 11px style table attributes columnalign left end attributes row cell straight x equals straight space rsinθcosϕ end cell row cell straight y equals straight space rsinθsinϕ end cell row cell straight z equals space rcosθ end cell row cell We space know space space space sin to the power of straight 2 straight A plus cos squared straight A equals 1 --- box enclose 1 end cell row cell rightwards double arrow space straight x to the power of straight 2 plus straight y to the power of straight 2 end cell row cell rightwards double arrow straight space straight r squared sin squared θcos squared straight ϕ plus straight r squared sin squared θsin squared straight ϕ end cell row cell rightwards double arrow straight space straight r squared sin squared straight theta left parenthesis cos squared straight ϕ plus sin squared straight ϕ right parenthesis end cell row cell rightwards double arrow straight space straight r squared sin squared straight theta --- box enclose 1 end cell row cell rightwards double arrow straight space straight r squared sin squared straight theta straight space equals space straight x to the power of straight 2 plus straight y to the power of straight 2 end cell row cell Add space straight z to the power of straight 2 space both space side end cell row cell rightwards double arrow space straight x to the power of straight 2 plus straight y to the power of straight 2 plus straight z to the power of straight 2 equals straight r squared sin squared straight theta plus straight r squared cos squared straight theta end cell row cell space space space space space = straight r squared left parenthesis sin squared straight theta plus cos squared straight theta right parenthesis end cell row cell rightwards double arrow space straight x to the power of straight 2 plus straight y to the power of straight 2 plus straight z to the power of straight 2 equals straight r squared end cell end table So comma space the space correct space option space is space left parenthesis straight a right parenthesis. end style$

Trigonometric Identities Exercise 11.58

Solution 20

$begin mathsize 11px style We space know table attributes columnalign left end attributes row cell text end text sin squared straight theta plus cos squared straight theta equals 1 end cell row cell rightwards double arrow cos squared straight theta equals 1 minus sin squared straight theta text --- end text box enclose 1 end cell row cell Given text s end text inθ plus sin squared straight theta text =1 --- end text box enclose 2 text end text end cell row cell text end text sinθ equals text 1- end text sin squared straight theta text end text end cell row cell text from end text box enclose 1 end cell row cell text end text box enclose sinθ equals cos squared straight theta end enclose text --- end text box enclose 3 end cell row cell To text find end text cos squared straight theta plus cos to the power of 4 straight theta end cell row cell text end text rightwards double arrow cos squared straight theta left parenthesis 1 plus cos squared straight theta right parenthesis end cell row cell text from end text box enclose 3 end cell row cell text end text rightwards double arrow sinθ left parenthesis 1 plus sinθ right parenthesis end cell row cell text end text rightwards double arrow sinθ plus sin squared straight theta end cell row cell text from end text box enclose 2 end cell row cell text end text rightwards double arrow text 1 end text end cell end table So comma space the space correct space option space is space left parenthesis straight b right parenthesis. end style$

Solution 21

$begin mathsize 11px style table attributes columnalign left end attributes row cell acosθ plus bsinθ equals straight m end cell row cell On text squaring end text end cell row cell text a end text to the power of text 2 end text end exponent cos squared straight theta plus straight b squared sin squared straight theta plus 2 absinθcosθ equals straight m squared text --- end text box enclose 1 end cell row cell asinθ minus bcosθ equals straight n end cell row cell On text squaring end text end cell row cell text a end text to the power of text 2 end text end exponent sin squared straight theta plus straight b squared cos squared straight theta minus 2 absinθcosθ equals straight n squared text --- end text box enclose 2 end cell row cell On text end text adding text end text box enclose 1 text & end text box enclose 2 text and knowing sin end text to the power of text 2 end text end exponent straight theta plus cos squared straight theta equals 1 end cell row cell rightwards double arrow text a end text to the power of text 2 end text end exponent left parenthesis sin squared straight theta plus cos squared straight theta right parenthesis plus straight b squared left parenthesis sin squared straight theta plus cos squared straight theta right parenthesis equals straight m squared plus straight n squared end cell row cell rightwards double arrow text a end text to the power of text 2 end text end exponent plus straight b to the power of text 2 end text end exponent equals straight m to the power of text 2 end text end exponent plus straight n to the power of text 2 end text end exponent end cell end table So comma space the space correct space option space is space left parenthesis straight d right parenthesis. end style$

Solution 22

$begin mathsize 11px style table attributes columnalign left end attributes row cell text end text sin squared straight A plus cos squared straight A equals 1 end cell row cell text end text sin squared straight A equals 1 minus cos squared straight A text --- end text box enclose 1 end cell row cell given text cosA + cos end text to the power of text 2 end text end exponent straight A equals 1 text --- end text box enclose 2 end cell row cell text cosA = 1-cos end text to the power of text 2 end text end exponent straight A end cell row cell text from end text box enclose text 1 end text end enclose end cell row cell text cosA end text equals text sin end text to the power of text 2 end text end exponent text A --- end text box enclose 3 end cell row cell To text find: s end text in squared straight A plus sin to the power of 4 straight A end cell row cell text end text rightwards double arrow sin squared straight A left parenthesis 1 plus sin squared straight A right parenthesis end cell row cell text end text from text end text box enclose 3 text end text end cell row cell text end text rightwards double arrow text cosA end text left parenthesis 1 plus text cosA end text right parenthesis end cell row cell text end text rightwards double arrow text cosA end text plus text cos end text to the power of text 2 end text end exponent text A end text end cell row cell text end text from text end text box enclose 2 end cell row cell text end text rightwards double arrow text 1 end text end cell end table So comma space the space correct space option space is space left parenthesis straight c right parenthesis. end style$

Solution 23

$begin mathsize 11px style We space know table attributes columnalign left end attributes row cell sin squared straight theta plus cos squared plus 1 text --- end text box enclose 1 end cell row cell 1 plus tan squared straight theta equals sec squared straight theta text --- end text box enclose 2 end cell row cell rightwards double arrow straight x squared over straight a squared text end text plus text end text straight y squared over straight b squared end cell row cell equals fraction numerator straight a squared sec squared θcos squared straight ϕ over denominator straight a squared end fraction text end text plus text end text fraction numerator straight b squared sec squared θsin squared straight ϕ over denominator straight b squared end fraction end cell row cell equals text end text sec squared θcos squared straight ϕ text end text plus text end text sec squared θsin squared straight ϕ end cell row cell equals text end text sec squared straight theta left parenthesis cos squared straight ϕ plus sin squared straight ϕ right parenthesis end cell row cell text from end text box enclose 1 text & end text box enclose 2 end cell row cell rightwards double arrow text end text left parenthesis 1 plus tan squared straight theta right parenthesis text --- end text box enclose 3 end cell row cell given text z = ctan end text straight theta end cell row cell rightwards double arrow text tan end text straight theta equals straight z over straight c text --- end text box enclose 4 end cell row cell rightwards double arrow straight x squared over straight a squared text end text plus text end text straight y squared over straight b squared equals 1 plus straight z squared over straight c squared end cell end table So comma space the space correct space option space is space left parenthesis straight d right parenthesis. end style$

Solution 24

$begin mathsize 11px style table attributes columnalign left end attributes row cell acosθ minus bsinθ equals straight c end cell row cell rightwards double arrow space space space On space squaring end cell row cell space space space space space space space straight a to the power of straight 2 cos squared straight theta plus straight b squared sin squared straight theta minus 2 abcosθsinθ equals straight c squared space space space minus negative negative box enclose 1 end cell row cell Let space space space asinθ plus bcosθ space equals space straight t end cell row cell space space space space space space space On space squaring end cell row cell space space rightwards double arrow space straight a to the power of straight 2 sin to the power of straight 2 straight theta space plus space straight b to the power of straight 2 cos to the power of straight 2 straight theta space plus space 2 absinθcosθ space equals space straight t to the power of straight 2 space space space minus negative negative box enclose 2 end cell row cell Adding straight space box enclose straight 1 space & space box enclose 2 comma space space we space get end cell row cell rightwards double arrow space straight a to the power of straight 2 left parenthesis sin to the power of straight 2 straight theta plus cos to the power of straight 2 straight theta right parenthesis space plus space straight b to the power of straight 2 left parenthesis sin to the power of straight 2 straight theta plus cos to the power of straight 2 straight theta right parenthesis equals space straight c to the power of straight 2 plus straight t squared space space space minus negative negative box enclose 3 end cell row cell rightwards double arrow space sin to the power of straight 2 straight theta plus cos to the power of straight 2 straight theta equals 1 space space space left parenthesis we space know right parenthesis end cell row cell space space space space space from space box enclose 3 end cell row cell space space space space straight a to the power of straight 2 plus straight b to the power of straight 2 equals straight c to the power of straight 2 plus straight t squared end cell row cell rightwards double arrow space straight t equals plus-or-minus space root index blank of straight a to the power of straight 2 plus straight b to the power of straight 2 minus straight c to the power of straight 2 end root end cell row cell Hence comma end cell row cell asinθ space plus space bcosθ space equals plus-or-minus space root index blank of straight a to the power of straight 2 plus straight b to the power of straight 2 minus straight c to the power of straight 2 end root end cell end table So comma space the space correct space option space is space left parenthesis straight b right parenthesis. end style$

Solution 25

$begin mathsize 11px style We space know table attributes columnalign left end attributes row cell text end text 1 text + tan end text to the power of text 2 end text end exponent straight a equals sec squared straight A end cell row cell rightwards double arrow sec squared straight A minus tan squared straight A equals 1 text --- end text box enclose 1 end cell row cell To text find: 9sec end text to the power of text 2 end text end exponent straight A minus 9 tan squared straight A end cell row cell text end text rightwards double arrow text 9 end text left parenthesis text sec end text to the power of text 2 end text end exponent straight A minus tan squared straight A right parenthesis end cell row cell text from end text box enclose 1 end cell row cell text end text rightwards double arrow text 9(1) = 9 end text end cell end table So comma space the space correct space option space is space left parenthesis straight b right parenthesis. end style$

Solution 26

$begin mathsize 11px style table attributes columnalign left end attributes row cell left parenthesis 1 plus sinθ over cosθ plus 1 over cosθ right parenthesis left parenthesis 1 plus cosθ over sinθ minus 1 over sinθ right parenthesis end cell row cell equals left parenthesis fraction numerator sinθ plus cosθ plus 1 over denominator cosθ end fraction right parenthesis left parenthesis fraction numerator sinθ plus cosθ minus 1 over denominator sinθ end fraction right parenthesis end cell row cell We space know end cell row cell left parenthesis straight a plus straight b right parenthesis left parenthesis straight a minus straight b right parenthesis plus straight a squared minus straight b squared --- box enclose 1 end cell row cell and sin to the power of straight 2 straight theta plus cos squared plus 1 space --- box enclose 2 end cell row cell from straight space box enclose 1 comma space we space get end cell row cell fraction numerator left parenthesis sinθ plus cosθ right parenthesis squared minus 1 over denominator sinθcosθ end fraction end cell row cell equals space fraction numerator sin to the power of straight 2 straight theta plus cos squared plus 2 sinθcosθ minus 1 over denominator sinθcosθ end fraction end cell row cell from straight space box enclose 2 end cell row cell rightwards double arrow straight space fraction numerator 1 plus 2 sinθcosθ minus 1 over denominator sinθcosθ end fraction end cell row cell rightwards double arrow 2 end cell row cell space So comma space the space correct space option space is space left parenthesis straight c right parenthesis. space space space end cell end table end style$

Solution 27

$begin mathsize 11px style table attributes columnalign left end attributes row cell text end text left parenthesis secA plus tanA right parenthesis left parenthesis 1 minus sinA right parenthesis end cell row cell equals text end text open parentheses 1 over cosA plus sinA over cosA close parentheses left parenthesis 1 minus sinA right parenthesis end cell row cell equals text end text fraction numerator left parenthesis 1 plus sinA right parenthesis left parenthesis 1 minus sinA right parenthesis over denominator cosA end fraction end cell row cell equals text end text fraction numerator 1 minus sin squared straight A over denominator cos squared end fraction end cell row cell equals text end text fraction numerator cos squared straight A over denominator cosA end fraction end cell row cell equals text cosA end text end cell end table So comma space the space correct space option space is space left parenthesis straight d right parenthesis. end style$

Solution 28

$begin mathsize 11px style We space know table attributes columnalign left end attributes row cell 1 plus tan squared straight A equals sec squared straight A equals fraction numerator 1 over denominator cos squared straight A end fraction text --- end text box enclose 1 end cell row cell 1 plus cot squared straight A equals cosec squared straight A equals fraction numerator 1 over denominator sin squared straight A end fraction text --- end text box enclose 1 end cell row cell To text find: end text fraction numerator 1 plus tan squared straight A over denominator 1 plus cot squared straight A end fraction end cell row cell from text end text box enclose 1 text & end text box enclose 2 end cell row cell text end text rightwards double arrow text end text fraction numerator fraction numerator 1 over denominator cos squared straight A end fraction over denominator fraction numerator 1 over denominator sin squared straight A end fraction end fraction text end text end cell row cell text end text rightwards double arrow text end text fraction numerator sin squared straight A over denominator cos squared straight A end fraction end cell row cell text end text rightwards double arrow text tan end text to the power of text 2 end text end exponent straight A end cell row cell text So, the correct option is (d). end text end cell end table end style$

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Class 10 RD SHARMA Solutions Maths Chapter 11 - Trigonometric Identities

Trigonometric Identities Exercise Ex. 11.1

Solution 1

Solution 2

Solution 3

Solution 4

Solution 5

Solution 6

Solution 7

Solution 8

Solution 9

Solution 10

Solution 11

Solution 12

Solution 13

Solution 14

Solution 15

Solution 16

Solution 17

Solution 18

Solution 19

Solution 20

Solution 21

Solution 22

Solution 23(i)

Solution 23(ii)

Solution 24

Solution 25

Solution 26

Solution 27

Solution 28

Solution 29

Solution 30

Solution 31

Solution 32

Solution 33

Solution 34

Solution 35

Solution 36

Solution 37. i

Solution 37. ii

Solution 38(i)

Solution 38(ii)

Solution 38(iii)

Solution 38(iv)

Solution 39

Solution 40

Solution 41

Solution 42

Solution 43

Solution 44

Solution 45

Solution 46

Solution 47 (i)

Solution 47 (ii)

Solution 47 (iii)

Solution 47(iv)

Solution 48

Solution 49

Solution 50

Solution 51

Solution 52

Solution 53

Solution 54

Solution 55. i

Solution 55. ii

Solution 56

Solution 57

Solution 58

Solution 59

Solution 60

Solution 61

Solution 62

Solution 63

Solution 64

Solution 65. i

Solution 65(ii)

Solution 66

Solution 67