which function is equivalent to $y = -cot(x)$?\n$y = -\tan(x)$\n$y = -\tan(x+\frac{pi}{2})$\n$y =…

Explanation:

Step1: Recall the co - function identity

We know that $\cot(x)=\frac{\cos(x)}{\sin(x)}$ and $\tan(x)=\frac{\sin(x)}{\cos(x)}$, and also the co - function identity $\tan\left(x +\frac{\pi}{2}\right)=-\cot(x)$. We start with the formula for the tangent of a sum: $\tan(A + B)=\frac{\tan(A)+\tan(B)}{1 - \tan(A)\tan(B)}$. When $A = x$ and $B=\frac{\pi}{2}$, $\tan\left(x+\frac{\pi}{2}\right)=\frac{\tan(x)+\tan\left(\frac{\pi}{2}\right)}{1-\tan(x)\tan\left(\frac{\pi}{2}\right)}$. In terms of sine and cosine, $\tan\left(x+\frac{\pi}{2}\right)=\frac{\sin\left(x+\frac{\pi}{2}\right)}{\cos\left(x+\frac{\pi}{2}\right)}$. Using the angle - addition formulas $\sin(A + B)=\sin(A)\cos(B)+\cos(A)\sin(B)$ and $\cos(A + B)=\cos(A)\cos(B)-\sin(A)\sin(B)$, we have $\sin\left(x+\frac{\pi}{2}\right)=\cos(x)$ and $\cos\left(x+\frac{\pi}{2}\right)=-\sin(x)$. So, $\tan\left(x+\frac{\pi}{2}\right)=-\frac{\cos(x)}{\sin(x)}=-\cot(x)$.

Answer:

$y = \tan\left(x+\frac{\pi}{2}\right)$