Question

an equivalent trigonometric expression for $\tan(x + \frac{\pi}{2})$ is
a) $\tan x$
b) $-\tan x$
c) $\cot x$
d) none of the above

Explanation:

Step1: Use the tangent addition formula

The tangent addition formula is $\tan(A + B)=\frac{\tan A+\tan B}{1-\tan A\tan B}$. Here $A = x$ and $B=\frac{\pi}{2}$. So, $\tan(x+\frac{\pi}{2})=\frac{\tan x+\tan\frac{\pi}{2}}{1 - \tan x\tan\frac{\pi}{2}}$. But $\tan\frac{\pi}{2}$ is undefined. We can also use the co - function identity. We know that $\tan\theta=\frac{\sin\theta}{\cos\theta}$. Then $\tan(x + \frac{\pi}{2})=\frac{\sin(x+\frac{\pi}{2})}{\cos(x+\frac{\pi}{2})}$.
Using the sum formulas: $\sin(A + B)=\sin A\cos B+\cos A\sin B$ and $\cos(A + B)=\cos A\cos B-\sin A\sin B$.
For $\sin(x+\frac{\pi}{2})=\sin x\cos\frac{\pi}{2}+\cos x\sin\frac{\pi}{2}=\cos x$ (since $\sin\frac{\pi}{2} = 1$ and $\cos\frac{\pi}{2}=0$).
For $\cos(x+\frac{\pi}{2})=\cos x\cos\frac{\pi}{2}-\sin x\sin\frac{\pi}{2}=-\sin x$ (since $\sin\frac{\pi}{2} = 1$ and $\cos\frac{\pi}{2}=0$).
So, $\tan(x+\frac{\pi}{2})=\frac{\cos x}{-\sin x}=-\cot x$.

Answer:

d) none of the above