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

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Solution

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

Hence, L.H.S. = R.H.S.

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