Question

Prove that sin (x+y) /sin (x−y) = tanx+tany/tanx−​tany

Answer 1

Solution:

We have to prove that:

$\rm \longrightarrow \dfrac{ \sin(x + y) }{ \sin(x - y) } = \dfrac{ \tan(x) + \tan(y) }{ \tan(x) - \tan(y) }$

Taking Left Hand Side, we get:

$\rm=\dfrac{ \sin(x + y) }{ \sin(x - y) }$

Can be written as:

$\rm=\dfrac{ \sin(x) \cos(y) + \cos(x) \sin(y) }{ \sin(x) \cos(y) - \cos(x) \sin(y) }$

Dividing both numerator and denominator by cos(x) cos(y), we get:

$\rm=\dfrac{ \dfrac{ \sin(x) \cos(y) + \cos(x) \sin(y) }{ \cos(x) \cos(y) }}{ \dfrac{\sin(x) \cos(y) - \cos(x) \sin(y) }{ \cos(x) \cos(y) }}$

$\rm=\dfrac{ \dfrac{ \sin(x) \cos(y)}{ \cos(x) \cos(y)} + \dfrac{ \cos(x) \sin(y) }{ \cos(x) \cos(y) } }{ \dfrac{ \sin(x) \cos(y)}{ \cos(x) \cos(y)} - \dfrac{ \cos(x) \sin(y) }{ \cos(x) \cos(y) } }$

$\rm=\dfrac{ \dfrac{ \sin(x)}{ \cos(x) } + \dfrac{\sin(y) }{\cos(y) } }{ \dfrac{ \sin(x)}{ \cos(x)} - \dfrac{\sin(y) }{\cos(y) } }$

$\rm = \dfrac{ \tan(x) + \tan(y) }{ \tan(x) - \tan(y) }$

$\rm = RHS$

Hence Proved.

Additional Information:

$\rm1. \: \sin(\alpha+\beta) = \sin(\alpha) \cos(\beta) + \cos(\alpha) \sin(\beta)$

$\rm2.\: \cos(\alpha+\beta) = \cos(\alpha) \cos(\beta) - \sin(\alpha) \sin(\beta)$

$\rm 3.\: \tan(\alpha+\beta) = \dfrac{ \tan( \alpha) + \tan( \beta ) }{1 - \tan( \alpha ) \tan( \beta ) }$

$\rm 4. \: \sin(\alpha - \beta) = \sin(\alpha) \cos(\beta) - \cos(\alpha) \sin(\beta)$

$\rm 5.\: \cos(\alpha - \beta) = \cos(\alpha) \cos(\beta) + \sin(\alpha) \sin(\beta)$

$\rm 6.\: \tan(\alpha - \beta) = \dfrac{ \tan( \alpha) - \tan( \beta ) }{1 + \tan( \alpha ) \tan( \beta ) }$