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CBSE Class 10 Answered

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Asked by barbysingh2006 | 30 Dec, 2022, 08:41: PM
answered-by-expert Expert Answer
Let C = cosθ  and S = sinθ
 
begin mathsize 14px style fraction numerator 1 over denominator s e c squared theta space minus space cos squared theta end fraction space equals space fraction numerator 1 over denominator begin display style fraction numerator 1 over denominator cos squared theta end fraction end style minus cos squared theta end fraction space equals space fraction numerator cos squared theta over denominator 1 space minus space cos to the power of 4 theta end fraction space equals space fraction numerator C squared over denominator 1 space minus space C to the power of 4 end fraction end style
 
 
begin mathsize 14px style fraction numerator 1 over denominator c o s e c squared theta space minus space sin squared theta end fraction space equals space fraction numerator 1 over denominator begin display style fraction numerator 1 over denominator sin squared theta end fraction end style minus sin squared theta end fraction space equals space fraction numerator sin squared theta over denominator 1 space minus space sin to the power of 4 theta end fraction space equals space fraction numerator S squared over denominator 1 space minus space S to the power of 4 end fraction end style
 
begin mathsize 14px style open square brackets fraction numerator 1 over denominator s e c squared theta space minus space cos squared theta end fraction space plus fraction numerator 1 over denominator c o s e c squared theta space minus space sin squared theta end fraction space close square brackets cross times sin squared theta space cos squared theta space equals space open square brackets fraction numerator C squared over denominator 1 space minus space C to the power of 4 end fraction plus fraction numerator S squared over denominator 1 minus S to the power of 4 end fraction close square brackets space cross times C squared S squared end style
 
begin mathsize 14px style R H S space equals space open square brackets fraction numerator C squared open parentheses 1 minus S to the power of 4 close parentheses space plus space S squared left parenthesis 1 minus C to the power of 4 right parenthesis over denominator open parentheses 1 space minus space C to the power of 4 close parentheses stretchy left parenthesis 1 space minus space S to the power of 4 stretchy right parenthesis end fraction close square brackets space cross times C squared S squared end style
 
begin mathsize 14px style R H S space equals space open square brackets fraction numerator C squared minus C squared S to the power of 4 plus S squared minus S squared C to the power of 4 over denominator open parentheses 1 space minus space C squared close parentheses open parentheses 1 plus C squared close parentheses stretchy left parenthesis 1 space minus space S squared stretchy right parenthesis open parentheses 1 plus S squared close parentheses end fraction close square brackets space cross times C squared S squared end style
 
Since (1-C2 ) = S2   ,  ( 1-S2) = C2  and  (S2,+ C2 )  = 1 ,  then we get
 
 
begin mathsize 14px style R H S space equals space open square brackets fraction numerator 1 space minus C squared S squared over denominator open parentheses 1 plus C squared close parentheses space open parentheses 1 plus S squared close parentheses end fraction close square brackets space space equals space open square brackets fraction numerator 1 space minus space C squared S squared over denominator 2 plus C squared S squared end fraction close square brackets end style
 
begin mathsize 14px style R H S space equals space space open square brackets 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 close square brackets end style
 

 


 
 
 
 
Answered by Thiyagarajan K | 30 Dec, 2022, 09:26: PM
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