Answer
Work on the left side of the identity using $\cot{\theta}=\frac{\cos\theta}{\sin\theta}$ and $\csc\theta=\frac{1}{\sin\theta}$..
Refer to the step-by-step part below for the complete proof.
Work Step by Step
We have to show that:
$(\csc\theta+\cot\theta)(\csc\theta-\cot\theta)=1$
By evaluating the left side we get:
$=\csc^2\theta+\csc\theta\cot\theta-\csc\theta\cot\theta-\cot^2\theta\\
=\csc^2\theta-\cot^2\theta$
By using $\csc\theta=\frac{1}{\sin\theta}$ and $\cot\theta=\frac{\cos\theta}{\sin\theta}$, the expression above simplifies to:
$=\dfrac{1}{\sin^2\theta}-\dfrac{\cos^2\theta}{\sin^2\theta}$
$=\dfrac{1-\cos^2\theta}{\sin^2\theta}$
$=\dfrac{\sin^2\theta}{\sin^2\theta}$
$=1$
Since LHS=RHS. then the identity's proof is complete.
