Question

prove that

$\sqrt{ \frac{1 + \cos( \alpha ) }{1 - \cos( \alpha ) } } = \csc( \alpha ) + \cot( \alpha )$
​

Answer 1

$<b>$LHS

$\bf\sqrt{ \frac{1 + \cos( \alpha ) }{1 - \cos( \alpha ) } }$

$\bf\sqrt{ \frac{1 + \cos( \alpha ) \times 1 + \cos( \alpha ) }{1 - \cos( \alpha ) \times 1 + \cos( \alpha ) } }$

$\bf \sqrt{ \frac{(1 + \cos( \alpha )) {}^{2} }{1 - \cos {}^{2} ( \alpha ) } }$

$\bf\sqrt{ \frac{(1 + \cos( \alpha )) {}^{2} }{ \sin {}^{2} ( \alpha ) } }$

$\bf\frac{1 + \cos( \alpha ) }{ \sin( \alpha ) }$

$\bf\frac{1}{ \sin( \alpha ) } + \frac{ \cos( \alpha ) }{ \sin( \alpha ) }$

$\bf \csc( \alpha ) + \cot( \alpha )$

RHS

Hence proved

Answer 2

Proof :

Now, L.H.S. = $\sqrt{\frac{1+cos\alpha}{1-cos\alpha}}$

Multiplying both the numerator and the denominator inside the square root by $(1+cos\alpha)$, we get

$\to \sqrt{\frac{(1+cos\alpha)(1+cos\alpha)}{(1-cos\alpha)(1+cos\alpha)}}$

$= \sqrt{\frac{(1+cos\alpha)(1+cos\alpha)}{1-cos^{2}\alpha}}$

$=\sqrt{\frac{(1+cos\alpha)^{2}}{sin^{2}\alpha}}$

         since , sin²α + cos²α = 1

$=\sqrt{\bigg(\frac{1+cos\alpha}{sin\alpha}\bigg)^{2}}$

$=\frac{1+cos\alpha}{sin\alpha}$

$=\frac{1}{sin\alpha}+\frac{cos\alpha}{sin\alpha}$

$=cosec\alpha+cot\alpha$ = R.H.S.

Hence, proved.