Question

$prove \: that$
•

$\sqrt{ \frac{1 + cos \: a}{1 - cos \: a } } = cosec \: a + cot \: a$
​

Answer 1

$\bold{\large{\underline{\underline{\sf{StEp\:by\:stEp\:explanation:}}}}}$

LHS =$\sf{ \sqrt{ \frac{1 + cos \: A}{1 - cos \: A} } }$

Multiply numerator and denominator with ( 1 + cos A )

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

$\sf { = \sqrt{ \frac{{1 + cos}^{2} }{ {1}^{2} + {cos \: }^{2}A } } }$

[ since ( a + b ) ( a - b ) = a²2 - b² ]

$\sf{ \sqrt{ \frac{ {1 + cos \: A}^{2} }{ {sin}^{2}A } } }$

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

$\sf{ \: = \frac{1}{sin \: A} + \frac{cos \: A}{sin \: A} }$

$\sf{ = cosec A + cot A }$

[HENCE PROVED]

Answer 2

$\huge{\boxed{\red{\boxed{\mathfrak{\pink{Solution}}}}}}$

$LHS =\sf{ \sqrt{ \frac{1 + cos \:A}{1 - cos \:A}}}$

Multiply numerator and denominator with ( 1 + cos A )

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

$\sf { = \sqrt{ \frac{{1 + cos}^{2} }{ {1}^{2} + {cos \:}^{2}A}}}$

[ since ( a + b ) ( a - b ) = a²2 - b² ]

$\sf{ = \frac{{1 + cos \:A}^{2}}{{sin}^{2}A}}$

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

$\sf{ \:= \frac{1}{sin \:A}+ \frac{cos \:A}{sin \:A}}$

$\sf{ = cosec A + cot A }$

[HENCE PROVED]