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Some Applications Of Trigonometry

Question

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Prove the following identity:
$fraction numerator cosecθ plus cotθ over denominator cosecθ minus cotθ end fraction space equals space 1 plus 2 cot squared straight theta plus 2 cosecθ. space cotθ$

Solution

straight L. straight H. straight S. space equals space fraction numerator cosecθ plus cotθ over denominator cosecθ minus cotθ end fraction space space space space space space space space space space space space space equals space fraction numerator cosecθ plus cotθ over denominator cosecθ minus cotθ end fraction cross times fraction numerator cosecθ plus cotθ over denominator cosecθ plus cotθ end fraction space space space space space space space space space space space space equals space fraction numerator left parenthesis cosecθ plus cotθ right parenthesis thin space left parenthesis cosecθ plus cotθ right parenthesis over denominator cosec squared straight theta minus cot squared straight theta end fraction space space space space space space space space space space space space equals space left parenthesis cosecθ plus cotθ right parenthesis squared space space space space space space space space space space space space equals space cosec squared straight theta plus cot squared straight theta plus 2 space cosecθ. space cotθ space space space space space space space space space space space space equals cot squared straight theta plus 1 plus cot squared straight theta plus 2 space cosecθ. space cotθ space space space space space space space space space space space space equals space 1 plus 2 cot squared straight theta plus 2 cosecθ. space cotθ space space space space space space space space space space space space space equals straight R. straight H. straight S.

L.H.S.= R.H.S.

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