Answer
$\cot(-\alpha)\cos(-\alpha)+\sin(-\alpha)=-\csc\alpha$
Work Step by Step
$\cot(-\alpha)\cos(-\alpha)+\sin(-\alpha)=-\csc\alpha$
On the left side of the equation, substitute $\cot(-\alpha)$ with $\dfrac{\cos(-\alpha)}{\sin(-\alpha)}$:
$\Big[\dfrac{\cos(-\alpha)}{\sin(-\alpha)}\Big]\cos(-\alpha)+\sin(-\alpha)=-\csc\alpha$
Since cosine is an even function, substitute $\cos(-\alpha)$ with $\cos(\alpha)$. Also, sine is an odd function, so substitute $\sin(-\alpha)$ with $-\sin(\alpha)$:
$\Big[\dfrac{\cos(\alpha)}{-\sin(\alpha)}\Big]\cos(\alpha)-\sin(\alpha)=-\csc\alpha$
$-\dfrac{\cos(\alpha)}{\sin(\alpha)}\cos(\alpha)-\sin(\alpha)=-\csc\alpha$
On the left side, factor out the minus sign and evaluate the sum of fractions and simplify:
$-\Big(\dfrac{\cos^{2}\alpha}{\sin\alpha}+\sin\alpha\Big)=-\csc\alpha$
$-\Big(\dfrac{\cos^{2}\alpha+\sin^{2}\alpha}{\sin\alpha}\Big)=-\csc\alpha$
Since $\cos^{2}\alpha+\sin^{2}=1$, the left side becomes:
$-\Big(\dfrac{1}{\sin\alpha}\Big)=-\csc\alpha$
The identity is proved, since $\dfrac{1}{\sin\alpha}=\csc\alpha$:
$-\csc\alpha=-\csc\alpha$
