Prove the following trigonometric identities.

In the given question, we need to prove $\frac{\cot A-\cos A}{\cot A+\cos A}=\frac{\operatorname{cosec} A-1}{\operatorname{cosec} A+1}$

Here, we will first solve the LHS.

Now, using $\cot \theta=\frac{\cos \theta}{\sin \theta}$, we get

$\frac{\cot A-\cos A}{\cot A+\cos A}=\frac{\left(\frac{\cos A}{\sin A}-\cos A\right)}{\left(\frac{\cos A}{\sin A}+\cos A\right)}$

$=\frac{\left(\frac{\cos A-\cos A \sin A}{\sin A}\right)}{\left(\frac{\cos A+\cos A \sin A}{\sin A}\right)}$

On further solving by taking the reciprocal of the denominator, we get,

$\frac{\left(\frac{\cos A-\cos A \sin A}{\sin A}\right)}{\left(\frac{\cos A+\cos A \sin A}{\sin A}\right)}=\left(\frac{\cos A-\cos A \sin A}{\sin A}\right)\left(\frac{\sin A}{\cos A+\cos A \sin A}\right)$

$=\left(\frac{\cos A-\cos A \sin A}{\cos A+\cos A \sin A}\right)$

Now, taking $\cos A \sin A$ common from both the numerator and the denominator, we get

$\left(\frac{\cos A-\cos A \sin A}{\cos A+\cos A \sin A}\right)=\frac{\cos A \sin A\left(\frac{1}{\sin A}-1\right)}{\cos A \sin A\left(\frac{1}{\sin A}+1\right)}$

$=\frac{\left(\frac{1}{\sin A}-1\right)}{\left(\frac{1}{\sin A}+1\right)}$

$=\frac{\operatorname{cosec} A-1}{\operatorname{cosec} A+1}$ $\left(\right.$ Using $\left.\frac{1}{\sin \theta}=\operatorname{cosec} \theta\right)$

Hence proved.