Question

√coseco-1/cosec +1= 1/seco+tano​

Answer 1

Solution :-

$\sf \sqrt{ \dfrac{cosec \theta - 1}{cosec \theta + 1} } = \dfrac{1}{sec \theta + tan \theta}$

Consider LHS

$\sf \sqrt{ \dfrac{cosec \theta - 1}{cosec \theta + 1} }$

$\sf = \sqrt{ \dfrac{ \dfrac{1}{sin \theta} - 1}{ \dfrac{1}{sin \theta} + 1} }$

$\sf = \sqrt{ \dfrac{ \dfrac{1 - sin \theta}{sin \theta}}{ \dfrac{1 + sin \theta}{sin \theta} } }$

$\sf = \sqrt{ \dfrac{ 1 - sin \theta}{1 + sin \theta}}$

On rationalising the denominator we get,

$\sf = \sqrt{ \dfrac{ 1 - sin \theta}{1 + sin \theta} \times \dfrac{1+ sin \theta}{1+ sin \theta} }$

$\sf = \sqrt{ \dfrac{(1-sin \theta)(1+sin \theta)}{( 1 + sin \theta)^2} }$

$\sf = \sqrt{ \dfrac{1^2 - sin^2 \theta}{(1 + sin \theta)^2}}$

[ Because (x + y)(x - y) = x² - y² ]

$\sf = \sqrt{ \dfrac{1 - sin^{2} \theta}{(1 + sin \theta)^2 }}$

$\sf = \sqrt{ \dfrac{cos^2 \theta}{(1 + sin \theta)^{2} } }$

[ Because 1 - sin²θ = cos² θ ]

$\sf = \sqrt{ \bigg( \dfrac{cos \theta}{1 + sin \theta } \bigg)^{2} }$

$\sf = \dfrac{cos \theta}{1 + sin \theta }$

Divide both numerator and denominator with cos θ

$\sf = \dfrac{ \dfrac{cos \theta}{cos \theta} }{ \dfrac{1 + sin \theta}{cos \theta } }$

$\sf = \dfrac{1}{ \dfrac{1}{cos \theta } + \dfrac{sin \theta}{cos \theta} }$

$\sf = \dfrac{1}{ sec \theta+ tan \theta}$

[ Because secθ = 1/cosθ and tanθ = sinθ/cosθ ]

= RHS

Hence proved.

Answer 2

$\huge\star{\red{\underline{\mathfrak{Answer :-}}}}$

Let us take the angle here as alpha instead of theta.

$\sqrt{ \dfrac{ \csc( \alpha ) - 1 }{ \csc( \alpha ) + 1 } } = \dfrac{1}{ \sec( \alpha ) + \tan( \alpha ) }$

Let's prove this...!!!

LHS :-

$\leadsto {\sqrt{ \dfrac{ \frac{1}{ \sin( \alpha ) } - 1 }{ \dfrac{1}{ \sin( \alpha ) } + 1} }}$

( cosec theta = 1/(Sin theta) )

$\leadsto { \sqrt \dfrac{ \dfrac{1 - \sin( \alpha ) }{ \sin( \alpha ) } }{ \dfrac{1 + \sin( \alpha ) }{ \sin( \alpha ) } }}$

$\leadsto{ \sqrt{ \dfrac{1 - \sin( \alpha ) }{ \sin( \alpha ) } \times \dfrac{ \sin( \alpha ) }{1 + \sin( \alpha ) } } }$

$\leadsto { \sqrt{ \dfrac{1 - \sin( \alpha ) }{1 + \sin( \alpha ) } }}$

Rationalizing the result :-

$\leadsto { \sqrt{ \dfrac{1 - \sin( \alpha ) }{1 + \sin( \alpha ) } } \times \sqrt{ \dfrac{1 - \sin( \alpha ) }{1 - \sin( \alpha ) }} }$

$\leadsto {\sqrt{ \dfrac{ {(1 - \sin( \alpha )) }^{2} }{ {1}^{2} - { \sin }^{2} \alpha } }}$

( cos^2 theta = 1- sin^2 theta )

$\leadsto { \sqrt{ \dfrac{ { \cos }^{2} \alpha }{ {(1 - \sin( \alpha )) }^{2} }} }$

$\leadsto { \dfrac{ \cos( \alpha ) }{1 - \sin( \alpha ) }}$

Dividing both numerator and denominator by cos alpha :-

$\leadsto { \dfrac{\dfrac {\cos(\alpha)}{\cos (\alpha)}}{\dfrac { 1+ \sin (\alpha)}{\cos (\alpha)}}}$

$\leadsto { \dfrac {1}{\dfrac {1+ \sin (\alpha)}{\cos(\alpha)}}}$

$\leadsto { \dfrac {1}{\dfrac {1}{\cos(\alpha)} + \dfrac {\sin(\alpha)}{\cos (\alpha)}}}$

( 1/(cos theta) = sec theta )

( sin theta / cos theta = tan theta )

$\leadsto {\dfrac{1}{ \sec( \alpha ) + \tan( \alpha ) }}$

$\therefore {Hence\:proved}$