Question

prove that sin(x+y)/sin(x-y) = tanx+Tany/tanx- tany​

Answer 1

Given that,

$\sf \dfrac{sin(x+y)}{sin(x-y)} = \dfrac{tanx+tany}{tanx- tany}$

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

we know that,

Sin (x + y) = sinx cosy + cosx siny and

Sin (x - y) = sinx cosy - cosx siny

$\sf = \dfrac{ sinx \: cosy + cosx \: siny}{sinx \: cosy - cosx \: siny}$

$\sf = \dfrac{ \dfrac{sinx \: cosy}{cosx \: cosy} + \dfrac{ cosx \: siny}{cosx \: cosy}}{ \dfrac{sinx \: cosy}{cosx \: cosy} - \dfrac{cosx \: siny}{cosx \: cosy}}$

$\sf = \dfrac{ \dfrac{sinx }{cosx } + \dfrac{ siny}{cosy }}{ \dfrac{sinx }{cosx } - \dfrac{ siny}{cosy}}$

we know that,

$\sf \dfrac{sinx}{cosx} = tanx$

$\sf = \dfrac{tanx + tany}{tanx - tany} = RHS$

LHS = RHS

So,

$\sf \dfrac{sin(x+y)}{sin(x-y)} = \dfrac{tanx+tany}{tanx- tany}$

Hence proved !