Question

tan o /(sec0-1) +tan0/(sec0+1)=2cosec​

Answer 1

Using LHS,

let theta = alpha (just for the sake of my comfort to type)

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



Hence proved.

Answer 2

Hence LHS=RHS

2cosec(0)=2cosec(0)

Hope it helps(•‿•)