Question

((Tan A+SecA)÷ (CosecA+CotA))×((TanA-SecA)÷CosecA-CotA)) = 2(TanA×CosecA-CotA×SecA)​

Answer 1

$\frac{ \sec(a) + \tan(a) }{ \cos \sec(a) + \cot(a) } = \frac{ \cos(a) - \cot(a) }{ \sec(a) - \tan(a) }$

LHS :-

$= \frac{ \sec(a) + \tan(a) }{ \cos \sec(a) + \cot(a) } \times \frac{ \sec(a) - \tan(a) }{ \sec(a) - \tan(a) } \times \frac{ \cos \sec(a) - \cot(a) }{ \cos \sec(a) - \cot(a) } \\ \\ = \frac{( \sec ^{2} (a) - { \tan }^{2} (a))( \cos \sec(a) - \cot(a) ) }{( \cos \sec ^{2} (a) - \cot^{2} (a))( \sec(a) - \tan(a)) } \\ \\ = \frac{(1)( \cos \sec(a) - \cot(a) ) }{(1)( \sec(a ) - \tan(a)) } \\ \\ = \frac{ \cos \sec(a) - \cot(a) }{ \sec(a) - \tan(a) }$

LHS = RHS

$\text{ Hence, proved }$$\large{\red{\ddot{\smile}}}$