Question

prove that; <br /><br />{1+cos A+sinA / 1+sin A-cosA}^2 = 1+cos A / 1-cos A here 1- cos A not equal to 0<br />​

Answer 1

Step-by-step explanation:

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 :-

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

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

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

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

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

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

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

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

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

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

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

Hence proved

Answer 2

Step-by-step explanation:

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