Prove the following identity:
$left parenthesis 1 plus tan squared straight theta right parenthesis plus open parentheses 1 plus fraction numerator 1 over denominator tan squared straight theta end fraction close parentheses space equals fraction numerator 1 over denominator sin squared straight theta minus sin to the power of 4 straight theta end fraction$

Solution

L.H.S. = $left parenthesis 1 plus tan squared straight theta right parenthesis plus open parentheses 1 plus fraction numerator 1 over denominator tan squared straight theta end fraction close parentheses$
$equals space left parenthesis 1 plus tan squared straight theta right parenthesis space plus space left parenthesis 1 plus cot squared straight theta right parenthesis space equals sec squared straight theta space plus space cosec squared straight theta space equals space fraction numerator 1 over denominator cos squared straight theta end fraction plus fraction numerator 1 over denominator sin squared straight theta end fraction space equals space fraction numerator sin squared straight theta plus cos squared straight theta over denominator cos squared straight theta. space sin squared straight theta end fraction equals space fraction numerator sin squared straight theta plus cos squared straight theta over denominator left parenthesis 1 minus sin squared straight theta right parenthesis space sin squared straight theta end fraction space equals space fraction numerator 1 over denominator sin squared straight theta minus sin to the power of 4 straight theta end fraction equals straight R. straight H. straight S.$
Hence, L.H.S. = R.H.S.

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Some Applications Of Trigonometry

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Choose the correct option and justify your choice:
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