Question

Tan+sec-1/tan+sec+1 = 1-sin/cos

Answer 1

Correct Question :

Tan A +Sec A - 1/Tan A + Sec A + 1 = 1 + Sin A /Cos A = Cos A / 1 - Sin A

Solution :

Taking LHS,

$\tt\dagger \: \: \: \: \: \dfrac{\tan A + \sec A - 1 }{\tan A - \sec A + 1 }$

$\tt \big\langle \: \: \: \: \: \: 1 + { \tan }^{2} \theta = { \sec }^{2} \theta.$

$\tt\longmapsto \dfrac{\tan A + \sec A - ( { \sec}^{2}A - { \tan }^{2}A ) }{\tan A - \sec A + 1 }$

$\tt \big\langle \: \: \: \: \: \: {a}^{2} - {b}^{2} = (a + b)(a - b)$

$\tt\longmapsto \dfrac{\tan A + \sec A - ( \tan A + \sec A )( \tan A - \sec A) }{\tan A - \sec A + 1 }$

Taking Common,

$\tt \longmapsto \dfrac{\tan A + \sec A [ 1 - (\sec A - \tan A ) ]}{\tan A - \sec A + 1 }$

$\tt\longmapsto \dfrac{\tan A + \sec A (1 - \sec A + \tan A ) }{\tan A - \sec A + 1 }$

$\tt\longmapsto \tan A + \sec A$

$\tt \longmapsto \dfrac{ \sin A }{ \cos A } + \dfrac{1}{ \cos A }$

$\tt \longmapsto \dfrac{ \sin A + 1}{ \cos A }$

$\rule{120}1$

Now,

Rationalizing the denominator.

$\tt \longmapsto \dfrac{ \sin A + 1}{ \cos A}$

$\tt \longmapsto \dfrac{ \sin A + 1}{ \cos A } \times \dfrac{ \cos A}{ \cos A }$

$\tt \longmapsto \dfrac{ \cos A(\sin A + 1)}{ {\cos}^2 A}$

$\tt \longmapsto \dfrac{ \cos A(\sin A + 1)}{ 1 - {\sin}^2 A}$

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

$\tt \longmapsto \dfrac{ \cos A }{ 1 - \sin A}$

LHS = RHS.

Hance Proved.