simplify tan(x + π/2). tan(x + π/2) = □

Explanation:

Step1: Use the tangent - addition formula

$\tan(A + B)=\frac{\tan A+\tan B}{1 - \tan A\tan B}$, where $A = x$ and $B=\frac{\pi}{2}$. But $\tan\frac{\pi}{2}$ is undefined. We use the co - function identity. We know that $\tan\left(x+\frac{\pi}{2}\right)=\frac{\sin\left(x + \frac{\pi}{2}\right)}{\cos\left(x+\frac{\pi}{2}\right)}$.

Step2: Apply the sum formulas for sine and cosine

$\sin(A + B)=\sin A\cos B+\cos A\sin B$ and $\cos(A + B)=\cos A\cos B-\sin A\sin B$. For $\sin\left(x+\frac{\pi}{2}\right)=\sin x\cos\frac{\pi}{2}+\cos x\sin\frac{\pi}{2}=\cos x$ (since $\cos\frac{\pi}{2} = 0$ and $\sin\frac{\pi}{2}=1$), and $\cos\left(x+\frac{\pi}{2}\right)=\cos x\cos\frac{\pi}{2}-\sin x\sin\frac{\pi}{2}=-\sin x$ (since $\cos\frac{\pi}{2} = 0$ and $\sin\frac{\pi}{2}=1$).

Step3: Calculate the tangent value

$\tan\left(x+\frac{\pi}{2}\right)=\frac{\sin\left(x+\frac{\pi}{2}\right)}{\cos\left(x+\frac{\pi}{2}\right)}=\frac{\cos x}{-\sin x}=-\cot x$.

Answer:

$-\cot x$