# Question #350e8

Dec 13, 2016

see below

#### Explanation:

$\sin \left(\pi - x\right) - \tan \left(\pi + x\right) = \sin x \frac{\cos x - 1}{\cos} x$

Use the formulas: $\sin \left(A - B\right) = \sin A \cos B - \cos A \sin B$

$\sin \left(A + B\right) = \sin A \cos B + \cos A \sin B$

$\cos \left(A - B\right) = \cos A \cos B + \sin A \sin B$

Left Hand Side: $= \sin \left(\pi - x\right) - \tan \left(\pi + x\right)$

$= \sin \left(\pi - x\right) - \frac{\sin \left(\pi + x\right)}{\cos} \left(\pi + x\right)$

$= \left(\sin \pi \cos x - \cos \pi \sin x\right) - \left(\frac{\sin \pi \cos x + \cos \pi \sin x}{\cos \pi \cos x - \sin \pi \sin x}\right)$

$= \left(0 \cdot \cos x - \left(- 1\right) \cdot \sin x\right) - \left(\frac{0 \cdot \cos x + \left(- 1\right) \sin x}{- 1 \cdot \cos x + 0 \cdot \sin x}\right)$

$= \sin x - \left(\frac{\cancel{-} \sin x}{\cancel{-} \cos x}\right)$

$= \sin x - \sin \frac{x}{\cos} x$

$= \frac{\sin x \cos x - \sin x}{\cos} x$

$= \sin x \frac{\cos x - 1}{\cos} x$

$\therefore =$ Right Handside

Dec 14, 2016

$L H S = \sin \left(\pi - x\right) - \tan \left(\pi + x\right)$

$= \sin x - \tan x$

$= \sin x - \sin \frac{x}{\cos} x$

$= \sin x \left(1 - \frac{1}{\cos} x\right)$

$= \sin x \left(\frac{\cos x - 1}{\cos} x\right) = R H S$

Proved