# IN any triangle ABC,sinA-cosB=cosC then angle B is?

## options given are $\frac{\pi}{2}$ $\frac{\pi}{3}$ $\frac{\pi}{4}$ $\frac{\pi}{6}$

Aug 10, 2018

$1 . B = \frac{\pi}{2}$

#### Explanation:

We know that,

$\left(1\right) \cos x + \cos y = 2 \cos \left(\frac{x + y}{2}\right) \cos \left(\frac{x - y}{2}\right)$

$\left(2\right) A + B + C = \pi = A = \pi - \left(B + C\right)$ $\to \left[\because \triangle A B C\right]$

$\left(3\right) \cos x - \cos y = - 2 \sin \left(\frac{x + y}{2}\right) \sin \left(\frac{x - y}{2}\right)$

Using $\left(1\right) \mathmr{and} \left(2\right)$

$\sin A = \cos B + \cos C$

$\therefore \sin \left[\pi - \left(B + C\right)\right] = 2 \cos \left(\frac{B + C}{2}\right) \cos \left(\frac{B - C}{2}\right)$

$\therefore \sin \left(B + C\right) = 2 \cos \left(\frac{B + C}{2}\right) \cos \left(\frac{B - C}{2}\right)$

Using $\sin 2 \theta = 2 \sin \theta \cos \theta ,$we get

$2 \sin \left(\frac{B + C}{2}\right) \cos \left(\frac{B + C}{2}\right) = 2 \cos \left(\frac{B + C}{2}\right) \cos \left(\frac{B - C}{2}\right)$

$\therefore \sin \left(\frac{B + C}{2}\right) = \cos \left(\frac{B - C}{2}\right) \to \left[\because \cos \left(\frac{B + C}{2}\right) \ne 0\right]$

$\cos \left[\frac{\pi}{2} - \frac{B + C}{2}\right]$=cos((B-C)/2)to[becausecos(pi/2- x)=sinx]

$\cos \left[\frac{\pi}{2} - \frac{B + C}{2}\right] - \cos \left(\frac{B - C}{2}\right) = 0$

Using $\left(3\right)$

$- 2 \sin \left(\frac{\frac{\pi}{2} - \frac{B + C}{2} + \frac{B - C}{2}}{2}\right) \sin \left(\frac{\frac{\pi}{2} - \frac{B + C}{2} - \frac{B - C}{2}}{2}\right) = 0$

Simplifying we get

$\sin \left(\frac{\pi}{4} - \frac{C}{2}\right) \sin \left(\frac{\pi}{4} - \frac{B}{2}\right) = 0$

$\sin \left(\frac{\pi}{4} - \frac{C}{2}\right) = 0 \mathmr{and} \sin \left(\frac{\pi}{4} - \frac{B}{2}\right) = 0$

$\implies \frac{\pi}{4} - \frac{C}{2} = 0 \mathmr{and} \frac{\pi}{4} - \frac{B}{2} = 0$

$\implies \frac{C}{2} = \frac{\pi}{4} \mathmr{and} \frac{B}{2} = \frac{\pi}{4}$

$\implies C = \frac{\pi}{2} \mathmr{and} B = \frac{\pi}{2}$

Hence , $B = \frac{\pi}{2}$