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write all the trigonometric identities and prove that (1/ sec^2 theta - cos ^2 theta ( + )1/ cosec^2 theta - sin^2 theta) sin^2 theta. cos ^2 theta = 1 - sin^2 theta. cos^2 theta / 2+ sin^2 theta . cos^2 theta by using trigonometric identities
Asked by abhishekpareekv | 10 Oct, 2020, 02:04: PM
answered-by-expert Expert Answer
To Prove
 
begin mathsize 14px style fraction numerator cos squared theta space sin squared theta over denominator s e c squared theta space minus space cos squared theta end fraction plus fraction numerator cos squared theta space sin squared theta over denominator cos e c squared theta space minus space sin squared theta end fraction space equals space fraction numerator 1 space minus space cos squared theta space sin squared theta over denominator 2 plus cos squared theta space sin squared theta end fraction end style  ..........................(1)
 
Let us use the relation secθ = 1/cosθ  and cosecθ = 1 / sinθ in LHS .  
 
LHS becomes
 
begin mathsize 14px style fraction numerator cos to the power of 4 theta space sin squared theta space over denominator 1 space minus space cos to the power of 4 theta end fraction plus space fraction numerator sin to the power of 4 theta space cos squared theta space over denominator 1 space minus space sin to the power of 4 theta end fraction equals space fraction numerator cos to the power of 4 theta space space over denominator 1 space plus space cos squared theta end fraction plus space fraction numerator sin to the power of 4 theta space space over denominator 1 space plus space sin squared theta end fraction space equals space fraction numerator cos to the power of 4 theta space plus space cos to the power of 4 theta space sin squared theta space plus sin to the power of 4 theta cos squared theta space plus space space sin to the power of 4 theta over denominator left parenthesis space 1 space plus space cos squared theta space right parenthesis space left parenthesis space 1 space plus space sin squared theta space right parenthesis end fraction end style ..................(2)
 
Let us consider the following relations
 
Error converting from MathML to accessible text. .................(3)
begin mathsize 14px style cos to the power of 4 theta space sin to the power of 4 theta space space plus space sin to the power of 4 theta space cos to the power of 4 theta space equals space cos squared theta space sin squared theta space left parenthesis space cos squared theta space plus space sin squared theta right parenthesis space equals space cos squared theta space sin squared theta end style .................(4)
begin mathsize 14px style left parenthesis space 1 space plus space cos squared theta space right parenthesis space left parenthesis space 1 space plus space sin squared theta space right parenthesis space equals space left parenthesis space 1 space plus space cos squared theta space plus space sin squared theta space plus space cos squared theta space sin squared theta space right parenthesis space equals space left parenthesis space 2 space plus space cos squared theta space sin squared theta space right parenthesis end style ............. (5)
By using relations (3), (4) and (5) in the right most part of relation-(2) , we rewrite relation (2) as
 
begin mathsize 14px style fraction numerator cos to the power of 4 theta space sin squared theta space over denominator 1 space minus space cos to the power of 4 theta end fraction plus space fraction numerator sin to the power of 4 theta space cos squared theta space over denominator 1 space minus space sin to the power of 4 theta end fraction equals space space fraction numerator 1 space minus space cos squared theta space sin squared theta over denominator 2 space plus space space cos squared theta space sin squared theta end fraction end style
Answered by Thiyagarajan K | 15 Oct, 2020, 12:58: PM

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