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RD Sharma Solution for Class 11 Science Mathematics Chapter 7 - Trigonometric Ratios of Compound Angles

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Our RD Sharma Textbook Solutions are considered extremely helpful for solving the tough questions which are asked in the CBSE Class 11 exam. TopperLearning Textbook Solutions are compiled by subject experts. Herein, you can find all the answers to the textbook questions for Chapter 7 - Trigonometric Ratios of Compound Angles.

All our solutions are created in accordance with the latest CBSE syllabus, and they are amended from time to time to provide the most relevant answers. Our free RD Sharma Solutions for CBSE Class 11 Mathematics will strengthen your fundamentals in Mathematics and will help you in your attempts to score more marks in the final examination. CBSE Class 11 students can refer to our solutions any time — while doing their homework and while preparing for the exam.

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Exercise/Page

RD Sharma Solution for Class 11 Science Mathematics Chapter 7 - Trigonometric Ratios of Compound Angles Page/Excercise 7.2

Solution 4

Solution 1(i)

Solution 1(ii)

Solution 1(iii)

Solution 1(iv)

Solution 2(i)

Solution 2(ii)

Solution 2(iii)

Solution 3

RD Sharma Solution for Class 11 Science Mathematics Chapter 7 - Trigonometric Ratios of Compound Angles Page/Excercise 7.1

Solution 19

fraction numerator sin space x. cos space y plus sin space y. cos space x over denominator sin space x. cos space y minus sin space y. cos space x end fraction equals fraction numerator a plus b over denominator a minus b end fraction
rightwards double arrow fraction numerator sin space x. cos space y plus sin space y. cos space x plus sin space x. cos space y minus sin space y. cos space x over denominator sin space x. cos space y plus sin space y. cos space x minus sin space x. cos space y plus sin space y. cos space x end fraction equals fraction numerator a plus b plus a minus b over denominator a plus b minus a plus b end fraction left square bracket U sin g space C o m p o n e n d o space a n d space D i v i d e n d o right square bracket
rightwards double arrow fraction numerator 2 sin space x. cos space y over denominator 2 sin space y. cos space x end fraction equals fraction numerator 2 a over denominator 2 b end fraction
rightwards double arrow fraction numerator tan space x over denominator tan space y end fraction equals a over b
H e n c e space P r o v e d

Solution 32

Syntax error from line 1 column 49 to line 1 column 73. Unexpected 'mstyle '.

Solution 33

tan space theta equals fraction numerator sin space alpha minus cos space alpha over denominator sin space alpha plus cos space alpha end fraction
rightwards double arrow tan space theta equals space fraction numerator tan space alpha minus 1 over denominator tan space alpha plus 1 end fraction left square bracket D i v i d i n g space b o t h space N u m e r a t o r space a n d space D e n o m i n a t o r space b y space cos space alpha right square bracket
rightwards double arrow tan space theta equals fraction numerator tan space alpha minus tan space begin display style pi over 4 end style over denominator 1 plus tan space begin display style pi over 4 end style. tan space alpha end fraction
rightwards double arrow tan space theta equals tan space open parentheses alpha minus pi over 4 close parentheses
rightwards double arrow theta equals alpha minus pi over 4 space space space left square bracket R e m o v i n g space tan space f r o m space b o t h space s i d e s right square bracket
rightwards double arrow cos space theta equals cos open parentheses alpha minus pi over 4 close parentheses space left square bracket T a k i n g space cos space o n space b o t h space s i d e s right square bracket
rightwards double arrow cos space theta equals cos space alpha. cos space pi over 4 plus sin space alpha. sin space pi over 4
rightwards double arrow cos space theta equals cos space alpha. fraction numerator 1 over denominator square root of 2 end fraction plus sin space alpha. fraction numerator 1 over denominator square root of 2 end fraction
rightwards double arrow cos space theta equals fraction numerator cos space alpha plus sin space alpha over denominator square root of 2 end fraction
rightwards double arrow square root of 2 space cos space theta equals sin space alpha plus cos space alpha
H e n c e space P r o v e d

Solution 34

R H S comma
fraction numerator p plus q over denominator 1 minus p q end fraction
equals fraction numerator tan left parenthesis A plus B right parenthesis plus tan thin space left parenthesis A minus B right parenthesis over denominator 1 minus tan left parenthesis A plus B right parenthesis. tan left parenthesis A minus B right parenthesis end fraction
equals fraction numerator begin display style fraction numerator tan space A plus tan thin space B over denominator 1 minus tan space A. tan space B end fraction end style plus begin display style fraction numerator tan space A minus tan space B over denominator 1 plus tan space A. tan space B end fraction end style over denominator 1 minus fraction numerator tan space A plus tan thin space B over denominator 1 minus tan space A. tan space B end fraction. fraction numerator tan space A minus tan space B over denominator 1 plus tan space A. tan space B end fraction end fraction
equals fraction numerator begin display style fraction numerator left parenthesis tan thin space A plus tan space B right parenthesis left parenthesis 1 plus tan space A. tan space B right parenthesis plus left parenthesis tan space A minus tan space B right parenthesis left parenthesis 1 minus tan space A. space tan thin space thin space B right parenthesis over denominator left parenthesis 1 minus tan space A. tan space B right parenthesis left parenthesis 1 plus tan space A. tan space B right parenthesis end fraction end style over denominator begin display style fraction numerator left parenthesis 1 minus tan space A. tan space B right parenthesis left parenthesis 1 plus tan space A. tan space B right parenthesis minus left parenthesis tan space A plus tan space B right parenthesis. left parenthesis tan space A minus tan space B right parenthesis over denominator left parenthesis 1 minus tan space A. space tan space B right parenthesis left parenthesis 1 plus tan space A. tan space B right parenthesis end fraction end style end fraction
equals fraction numerator tan space A plus tan space B plus tan squared A. tan space B plus tan space A. tan squared B plus tan space A minus tan thin space B minus tan thin space squared A. tan space B plus tan space A. tan squared B over denominator 1 minus tan squared space A. tan squared space B minus tan squared space A plus tan squared space B end fraction
equals fraction numerator 2 space tan space A plus 2 space tan space A. tan squared B over denominator left parenthesis 1 minus tan squared A right parenthesis left parenthesis 1 plus tan squared B right parenthesis end fraction equals fraction numerator 2 space tan space A left parenthesis 1 plus tan squared space B right parenthesis over denominator left parenthesis 1 minus tan squared space A right parenthesis left parenthesis 1 plus tan squared space B right parenthesis end fraction equals fraction numerator 2 space tan space A over denominator 1 minus tan squared space A end fraction equals tan space 2 A equals L H S
H e n c e space P r o v e d

Solution 1(i)

Solution 1(ii)

Solution 1(iii)

Solution 1(iv)

Solution 2(a)

Solution 2(b)

Solution 3

Solution 4

Solution 5

Solution 6(i)

Solution 6(ii)

Solution 7

Solution 8(i)

Solution 8(ii)

Solution 8(iii)

Solution 9

Solution 10

Solution 11(i)

Solution 11(ii)

Solution 11(iii)

Solution 12(i)

  

Solution 12(iii)

 

  

Solution 12(ii)

  

Solution 13

Solution 14(i)

Solution 14(ii)

Solution 15(i)

Solution 15(ii)

Solution 16(i)

Solution 16(ii)

Solution 16(iii)

Solution 16(iv)

Solution 16(v)

Solution 16(vi)

Solution 17(i)

Solution 17(ii)

Solution 17(iii)

Solution 17(iv)

Solution 18

Solution 20

Solution 21

Solution 22

Solution 23

Solution 24

Solution 25

Solution 26

Solution 27

Solution 28(i)

Solution 28(ii)

Solution 29(i)

Solution 29(ii)

Solution 29(iii)

Solution 30

Solution 31

RD Sharma Solution for Class 11 Science Mathematics Chapter 7 - Trigonometric Ratios of Compound Angles Page/Excercise 7VSAQ

Solution 1

Solution 2

Solution 3

Solution 4

Solution 5

Solution 6

Solution 7

Solution 8

Solution 9

Solution 10

Solution 11

Solution 12

TopperLearning provides step-by-step solutions for each question in each chapter in the RD Sharma textbook for class 11. Access the CBSE Class 11 Mathematics Chapter 7 - Trigonometric Ratios of Compound Angles for free. The textbook questions have been solved by our subject matter experts to help you understand how to answer them. Our RD Sharma Textbook Solutions will help you to study and revise, and you can easily clear your fundamentals of Chapter 7 - Trigonometric Ratios of Compound Angles.

Text Book Solutions

CBSE XI Science - Mathematics

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