Question

Prove the following identities, where the angles involved are acute angles for which the expressions are defined.
$(\csc \theta -\cot \theta {)}^2=\frac{1-\cos \theta }{1+\cos \theta }$.

Answer 1

Consider the $LHS$ of the expression.

$\Rightarrow {(\csc \theta -\cot \theta )}^2$

$\Rightarrow {(\frac{1}{\sin \theta }-\frac{\cos \theta }{\sin \theta })}^2$

$\Rightarrow {\left(\frac{1-\cos \theta }{\sin \theta }\right)}^2$

$\Rightarrow \frac{{(1-\cos \theta )}^2}{{\sin }^2\theta }$

$\Rightarrow \frac{{(1-\cos \theta )}^2}{1-{\cos }^2\theta }$

$\Rightarrow \frac{(1-\cos \theta )(1-\cos \theta )}{(1+\cos \theta )(1-\cos \theta )}$

$\Rightarrow \frac{1-\cos \theta }{1+\cos \theta }=RHS$

Hence, proved.