Question

prove that : (sin A +cos A/sinA - cosA) +(sinA - cos A / sin A +cos A )=2 /1 - 2 cos^2A​

Answer 1

To Prove :-

$\sf \dfrac{ \sin(a) + \cos(a) }{ \sin(a) - \cos(a) } + \dfrac{ \sin(a) - \cos(a) }{ \sin(a) + \cos(a) } = \dfrac{2}{1 - { \cos(a) }^{2} }$

Solution :-

$\sf \implies \dfrac{ \sin(a) + \cos(a) }{ \sin(a) - \cos(a) } + \dfrac{ \sin(a) - \cos(a) }{ \sin(a) + \cos(a) } = \dfrac{2}{1 - { \cos(a) }^{2} }$

Taking LHS :-

$\sf \implies \dfrac{ \sin(a) + \cos(a) }{ \sin(a) - \cos(a) } + \dfrac{ \sin(a) - \cos(a) }{ \sin(a) + \cos(a) }$

$\sf \implies \dfrac{( \sin(a) + \cos(a) {)}^{2} + ( { \sin(a) - \cos(a) })^{2} }{ \sin(a)^{2} - \cos(a)^{2} }$

$\sf \implies \: \dfrac{ { {\sin(a)}^{2} + { \cos(a) }^{2} + 2 \sin(a) \cos(a) + { \sin(a) }^{2} + { \cos(a) }^{2} - 2 \sin(a) \cos(a) }^{2} }{ { \sin(a) }^{2} - { \cos(a) }^{2} }$

$\sf \implies \: we \: know \: that \: { \sin(a) }^{2} + { \cos(a) }^{2} = 1$

$\sf \implies \: \dfrac{ { 1 + 2 \sin(a) \cos(a) + 1 - 2 \sin(a) \cos(a) }}{ { \sin(a) }^{2} - { \cos(a) }^{2} }$

$\sf \implies \: \dfrac{ { 2 + 2 \sin(a) \cos(a) - 2 \sin(a) \cos(a) } }{ { \sin(a) }^{2} - { \cos(a) }^{2} }$

$\sf \implies \: \dfrac{ { 2 + \cancel{2 \sin(a) \cos(a) } \cancel {- 2 \sin(a) \cos(a) } } }{ { \sin(a) }^{2} - { \cos(a) }^{2} }$

$\sf \implies \: \dfrac{ { 2 }}{ { \sin(a) }^{2} - { \cos(a) }^{2} }$

$\sf \implies \: we \: know \: that \: { \sin(a) }^{2} =1 - { \cos(a) }^{2}$

$\sf \implies \: \dfrac{ { 2 }}{ 1 - { \cos(a) }^{2} - { \cos(a) }^{2} }$

$\sf \implies \: \dfrac{ { 2 }}{ 1 - 2 { \cos(a) }^{2} }$

Hence proved