Prove the following:$ \frac{\tan \left(90^{\circ}-A\right) \cot A}{\operatorname{cosec}^{2} A}-\cos ^{2} A=0 $

To do:

We have to prove that \( \frac{\tan \left(90^{\circ}-A\right) \cot A}{\operatorname{cosec}^{2} A}-\cos ^{2} A=0 \).

Solution:  

We know that,

$tan\ (90^{\circ}- \theta) = cot\ \theta$

$sin\ \theta=\frac{1}{cosec\ \theta}$

$\cot \theta=\frac{\cos\ \theta}{\sin\ \theta}$

Therefore,

$\frac{\tan \left(90^{\circ}-A\right) \cot A}{\operatorname{cosec}^{2} A}-\cos ^{2} A=\frac{\cot A \times \cot A}{\operatorname{cosec}^{2} A}-\cos ^{2} A$

$=\cot^2 A \times \frac{1}{\operatorname {cosec}^{2} A}-\cos ^{2} A$

$=\frac{cos^{2} A}{sin^{2} A}\times sin^{2} A-cos^{2} A$

$=cos^2 A-cos^2 A$

$=0$

Hence proved.