Question

sin A - sin B/cos A + cos B+
cos A - cos B/sin A+ sin B=0
​

Answer 1

we know that,

sin²A + cos²A = 1

RHS

= 0

LHS

$= \frac{sinx - siny}{cosx + cosy} + \frac{cosx - cosy}{sinx + siny} \\ = \frac{(sinx - siny)(sinx + siny) + (cosx - cosy)(cosx + cosy)}{(cosx + cosy)(sinx + siny)} \\ = \frac{ {sin}^{2}x - {sin}^{2}y + {cos}^{2} x - {cos}^{2} y}{(cosx + cosy)(sinx + siny)} \\ = \frac{ {sin}^{2}x + {cos}^{2} x - {sin}^{2} y - {cos}^{2} y}{(cosx + cosy)(sinx + siny)} \\ = \frac{( {sin}^{2} x + {cos}^{2}x) - ( {sin}^{2}y + {cos}^{2}y) }{(cosx + cosy)(sinx + siny) } \\ = \frac{1 - 1}{(cosx + cosy)(sinx + siny)} \\ = \frac{0}{(cosx + cosy)(sinx + siny)} \\ = 0$

= RHS

Since, LHS = RHS

$hence \\ \frac{sinx - siny}{cosx + cosy} + \frac{cosx - cosy}{sinx + siny} = 0$

... Hence Proved!