Prove that:$ \sin ^{2} A \cos ^{2} B-\cos ^{2} A \sin ^{2} B=\sin ^{2} A-\sin ^{2} B $

To do:

We have to prove that \( \sin ^{2} A \cos ^{2} B-\cos ^{2} A \sin ^{2} B=\sin ^{2} A-\sin ^{2} B \).

Solution:

We know that,

$\sin^2 A+\cos^2 A=1$

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

$\sec^2 A-\tan^2 A=1$

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

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

$\operatorname{cosec} A=\frac{1}{\sin A}$

$\sec A=\frac{1}{\cos A}$

Therefore,

$\sin ^{2} A \cos ^{2} B-\cos ^{2} A \sin ^{2} B=\sin ^{2} A (1-\sin ^{2} B)-(1-\sin ^{2} A) \sin ^{2} B$

$=\sin ^{2} A -\sin ^{2} A\sin ^{2} B-\sin ^{2} B+\sin ^{2} A \sin ^{2} B$

$=\sin ^{2} A-\sin ^{2} B$

Hence proved.