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

To do:

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

Solution:  

We know that,

$sin\ (90^{\circ}- \theta) = cos\ \theta$

$cos\ (90^{\circ}- \theta) = sin\ \theta$

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

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

Let us consider LHS,

$\frac{\cos \left(90^{\circ}-A\right) \sin \left(90^{\circ}-A\right)}{\tan \left(90^{\circ}-A\right)}=\frac{\sin A \cos A}{\cot A}$

$=\frac{\sin A \cos A}{\frac{\cos A}{\sin A}}$

$=\frac{\sin A \cos A \times \sin A}{\cos A}$

$=\sin A \times \sin A$

$=\sin^2 A$

$=$ RHS

Hence proved.

Tutorialspoint

Simply Easy Learning

Updated on: 10-Oct-2022

24 Views

Kickstart Your Career

Get certified by completing the course

Get Started