Question

1+ sin A-COS A/1+sinA+cosA= √1-cosA/1+cosA
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Answer 1

let's solve R.H.S first

$\displaystyle \implies \sqrt{ \dfrac{1 - \cos(x) }{1 + \cos(x) } \times \dfrac{1 - \cos(x) }{1 - \cos(x) } } \\ \implies \sqrt{ \frac{ {(1 - \cos \: x })^{2} }{ \sin ^{2} (x) } } = \frac{1 - \cos(x) }{ \sin(x) } \\ \implies\csc(x) - \cot(x)$

Now, let's find the value of L.H.S

$\implies\ \dfrac{1 + \sin(x) - \cos(x) }{1 + \sin(x) + \cos(x) } \\ \\ \implies\frac{ \dfrac{1 + \sin(x) - \cos(x)}{ \sin(x) } }{ \dfrac{1 + \sin(x) + \cos(x)}{ \sin(x) } } \\ \\ \implies\dfrac{\csc(x ) + 1 - \cot(x) }{ \csc(x) + 1 + \cot(x) } \\ \\ \implies\frac{ \csc( x) - \cot(x) - ( \csc ^{2} (x) - \cot ^{2} (x) ) }{\csc(x) + 1 + \cot(x) } \\ \\ \implies \frac{ \csc(x) - \cot(x) - ( \csc(x) + \cot(x) )( \csc(x) - \cot(x) ) }{\csc(x) + 1 + \cot(x) } \\ \\ \implies\frac{\csc(x) - \cot(x)( \cancel{\csc(x) + 1 + \cot(x) )} }{ \cancel{\csc(x) + 1 + \cot(x) } } \\ \\ \implies\csc(x) - \cot(x)$

$\sf\purple{since, L.H.S=R.H.S}$

$\sf\pink{Hence, proved}$

hope it's beneficial :D