Question

Only a genius can solve it !!

Answer 1

$\text{L.H.S}\\\\=\dfrac{\cos A}{1-\sin A} + \dfrac{\sin A}{1-\cos A} +1\\\\\\=\dfrac{\cos A(1-\cos A) + \sin A(1-\sin A) + (1-\sin A)(1 - \cos A)}{(1-\sin A)(1 -\cos A)}\\\\\\=\dfrac{\cos A - \cos^2 A + \sin A - \sin^2 A + 1 - \cos A - \sin A + \sin A \cos A}{(1 -\sin A)(1 - \cos A)}\\\\\\=\dfrac{-(\sin^2 A + \cos^2A) +1 + \sin A \cos A}{(1 - \sin A)(1 - \cos A)}\\\\\\=\dfrac{-1 + 1 + \sin A \cos A }{(1 - \sin A)((1 - \cos A)}\\\\\\=\dfrac{0+ \sin A \cos A}{(1 - \sin A)(1 - \cos A)}$

$=\dfrac{\sin A \cos A}{(1 - \sin A)(1 - \cos A)}\\\\\\=\text{R.H.S}\\\\\text{Proved.}$

Answer 2

Given:-

$\\ \sf \implies \frac{ \cos(A) }{1 - \sin(A) } + \frac{ \sin(A) }{1 - \cos(A) } + 1$

To Prove :-

$\\ \sf \implies \frac{ \sin(A) \cos(A) }{(1 - \sin(A)(1 - \cos(A)}$

Solution:-

$\\ \sf \implies \: LHS = \: \frac{ \cos(A) }{1 - \sin(A) } + \frac{ \sin(A) }{1 - \cos(A) } + 1$

$\\ \sf \implies \: LHS = \: \frac{ \cos( 1 - \cos A ) + \sin A(1 -\sin A ) } {(1 -\sin A)( 1 - \cos A)}+ 1$

$\\ \sf \implies \: LHS = \: \frac{\cos A - \cos {}^{2} A + \sin A - \sin {}^{2} A + (1 -\sin A)( 1 - \cos A)} {(1 -\sin A)( 1 - \cos A)}$

$\\ \sf \implies \: LHS = \: \frac{\cos A + \sin A - ( \cos {}^{2} A - \sin {}^{2} A )+ 1 - \cos A - \sin A + \cos A \sin A } {(1 -\sin A)( 1 - \cos A)}$

$\\ \sf \implies \: LHS = \: \frac{\cos A + \sin A - 1+ 1 - \cos A - \sin A + \cos A \sin A } {(1 -\sin A)( 1 - \cos A)}$

$\\ \sf \implies \: LHS = \: \frac { \cancel{\cos A } + \sin A - \cancel {1}+ \cancel {1 }- \cos A - \cancel {\sin A } +\cancel {\cos A } \: \: \cancel {\sin A } } {(1 -\sin A)( 1 - \cos A)}$

$\\ \sf \implies \frac{ \sin A \cos A }{(1 - \sin A)(1 - \cos A)}$

$\\ \sf \implies \: LHS = RHS$

$\\ \sf \implies \frac{ \cos(A) }{1 - \sin(A) } + \frac{ \sin(A) }{1 - \cos(A) } + 1 =\frac{ \sin(A) \cos(A) }{(1 - \sin(A)(1 - \cos(A)}$

Hence Proved !!