Answer
$\tan\theta+\cot\theta=\sec\theta\csc\theta$
Work Step by Step
$\tan\theta+\cot\theta=\sec\theta\csc\theta$
On the left side of the equation, substitute $\tan\theta$ with $\dfrac{\sin\theta}{\cos\theta}$ and $\cot\theta$ with $\dfrac{\cos\theta}{\sin\theta}$:
$\dfrac{\sin\theta}{\cos\theta}+\dfrac{\cos\theta}{\sin\theta}=\sec\theta\csc\theta$
Evaluate the sum of fractions:
$\dfrac{\sin^{2}\theta+\cos^{2}\theta}{\sin\theta\cos\theta}=\sec\theta\csc\theta$
Since $\sin^{2}\theta+\cos^{2}\theta=1$, the expression on the left becomes:
$\dfrac{1}{\sin\theta\cos\theta}=\sec\theta\csc\theta$
The fraction on the left, can be rewritten like this:
$\Big(\dfrac{1}{\sin\theta}\Big)\Big(\dfrac{1}{\cos\theta}\Big)=\sec\theta\csc\theta$
Since $\sec\theta=\dfrac{1}{\cos\theta}$ and $\csc\theta=\dfrac{1}{\sin\theta}$, the identity is proved:
$\sec\theta\csc\theta=\sec\theta\csc\theta$
