Question

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Answer 1

Answer:

$\frac{tan\alpha}{sec\alpha -1} + \frac{tan\alpha}{sec\alpha +1} =2cosec\alpha \\\\LHS\\\\\frac{tan\alpha}{sec\alpha -1} + \frac{tan\alpha}{sec\alpha +1}\\\\=\frac{tan\alpha(sec\alpha +1)+tan\alpha(sec\alpha -1)}{(sec\alpha -1)(sec\alpha +1)} \\\\=\frac{tan\alpha\sec\alpha +tan\alpha +tan\alpha\sec\alpha -tan\alpha }{sec^{2} \alpha -1} \\\\=\frac{2tan\alpha\sec\alpha}{tan^{2}\alpha } \\\\= \frac{2sec\alpha}{tan\alpha } \\\\=\frac{2cos\alpha }{cos\alpha sin\alpha } \\\\=\frac{2}{sin\alpha }$

$=2cosec\alpha$

therefore LHS=RHS