Answer
$\dfrac{\tan v-\cot v}{\tan^{2}v-\cot^{2}v}=\sin v\cos v$
Work Step by Step
$\dfrac{\tan v-\cot v}{\tan^{2}v-\cot^{2}v}=\sin v\cos v$
Factor the denominator on the left side and simplify:
$\dfrac{\tan v-\cot v}{(\tan v-\cot v)(\tan v+\cot v)}=\sin v\cos v$
$\dfrac{1}{\tan v+\cot v}=\sin v\cos v$
Replace $\tan v$ by $\dfrac{\sin v}{\cos v}$ and $\cot v$ by $\dfrac{\cos v}{\sin v}$:
$\dfrac{1}{\dfrac{\sin v}{\cos v}+\dfrac{\cos v}{\sin v}}=\sin v\cos v$
Evaluate the sum on the denominator and simplify:
$\dfrac{1}{\dfrac{\sin^{2}v+\cos^{2}v}{\sin v\cos v}}=\sin v\cos v$
$\dfrac{\sin v\cos v}{\sin^{2}v+\cos^{2}v}=\sin v\cos v$
Since $\sin^{2}v+\cos^{2}v=1$, the identity is proved.
$\sin v\cos v=\sin v\cos v$
