prove the identity.\n\frac{cos(\frac{pi}{2}+x)}{cos(pi + x)}=\tan x

Explanation:

Step1: Use the sum - of - angles formula for cosine

We know that $\cos(A + B)=\cos A\cos B-\sin A\sin B$. For $\cos(\frac{\pi}{2}+x)$, we have $\cos\frac{\pi}{2}\cos x-\sin\frac{\pi}{2}\sin x=0\times\cos x - 1\times\sin x=-\sin x$. For $\cos(\pi + x)$, we have $\cos\pi\cos x-\sin\pi\sin x=- 1\times\cos x-0\times\sin x=-\cos x$.

Step2: Substitute the results into the left - hand side of the identity

The left - hand side $\frac{\cos(\frac{\pi}{2}+x)}{\cos(\pi + x)}=\frac{-\sin x}{-\cos x}$.

Step3: Simplify the fraction

$\frac{-\sin x}{-\cos x}=\frac{\sin x}{\cos x}$. Since $\tan x=\frac{\sin x}{\cos x}$, the left - hand side is equal to the right - hand side.

Answer:

The identity $\frac{\cos(\frac{\pi}{2}+x)}{\cos(\pi + x)}=\tan x$ is proved.