Question

5. prove
1 + Sec A/sec A= sin^2 A/ 1- cos A​

Answer 1

Given :

$\tt\: \frac{1 + \sec \: A }{ \sec \: A } = \frac{ { \sin}^{2} A}{1 - \cos \: A }$

Solution :

Taking LHS,

$\tt\longmapsto \frac{1 + \sec \: A }{ \sec \: A }$

$\tt\longmapsto \frac{1 + \frac{1}{ \cos A } }{ \frac{1}{ \cos A } }$

$\tt\longmapsto \frac{ \cos A + 1}{1}$

$\tt\longmapsto { \cos A + 1 . }$

$\rule{90}1$

Taking RHS,

$\tt\longmapsto \frac{1 - { \cos A }^{2} }{1 - \cos A }$

$\tt \leadsto \fbox {{ \sin }^{2} \theta + { \cos}^{2} \theta = 1.}$

$\tt \longmapsto \frac{(1 + \cos A )(1 - \cos A ) }{1 - \cos A }$

$\tt\longmapsto1 + \cos A .$

Hence Proved, LHS = RHS.

$\rule{200}3$

Some Identities

• Sin²theta + Cos² theta = 1.
• 1 + tan² theta = Sec² theta.
• 1 + Cot² theta = Cosec² theta.