Question

prove that 1+cos alpha +sin alpha ÷ 1+cos alpha - sin alpha = 1+sin alpha ÷ cos alpha ​​

Answer 1

Required Answer:-

Given To Prove:

• (1 + cos α + sin α)/(1 + cos α - sin α) = (1 + sin α)/cos α

Proof:

Taking LHS,

$\sf \dfrac{1 + \cos( \alpha ) + \sin( \alpha ) }{1 + \cos( \alpha ) - \sin( \alpha ) }$

Multiplying numerator and denominator by cos α, we get,

$\sf = \dfrac{ \cos( \alpha ) \{1 + \cos( \alpha ) + \sin( \alpha ) \} }{ \cos( \alpha ) \{1 + \cos( \alpha ) - \sin( \alpha ) \} }$

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

We know that,

$\sf \implies { \sin}^{2}( \alpha ) + \cos^{2} ( \alpha ) = 1$

$\sf \implies \cos^{2} ( \alpha ) = 1 - { \sin}^{2}( \alpha )$

Putting the value of cos²α here, we get,

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

Now, simplify 1 - sin²α using identity a² - b² = (a + b)(a - b).

$\sf = \dfrac{( 1 + { \sin} ( \alpha )(1 - \sin( \alpha )) + \cos( \alpha ) \{1+ \sin( \alpha ) \} }{ \cos( \alpha ) \{1 + \cos( \alpha ) - \sin( \alpha ) \} }$

Taking 1 + sin α as common, we get,

$\sf = \dfrac{1 + { \sin} ( \alpha ) \{(1 - \sin( \alpha ) + \cos( \alpha ) \}}{ \cos( \alpha ) \{1 + \cos( \alpha ) - \sin( \alpha ) \} }$

$\sf = \dfrac{(1 + { \sin} ( \alpha ) ) \cancel{\{(1+ \cos( \alpha ) - \sin( \alpha ) \}}}{ \cos( \alpha ) \cancel{ \{1 + \cos( \alpha ) - \sin( \alpha ) \} }}$

$\sf = \dfrac{1 + \sin( \alpha ) }{ \cos( \alpha ) }$

= RHS (Proved)

Therefore,

$\sf \leadsto \dfrac{1 + \cos( \alpha ) + \sin( \alpha ) }{1 + \cos( \alpha ) - \sin( \alpha ) } = \dfrac{1 + \sin( \alpha ) }{ \cos( \alpha ) }$

Basic Formula Used:

• sin²α + cos²α = 1
• a² - b² = (a + b)(a - b)