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Prove that:$ \tan ^{2} A \sec ^{2} B-\sec ^{2} A \tan ^{2} B=\tan ^{2} A-\tan ^{2} B $
To do:
We have to prove that \( \tan ^{2} A \sec ^{2} B-\sec ^{2} A \tan ^{2} B=\tan ^{2} A-\tan ^{2} B \).
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,
$\tan ^{2} A \sec ^{2} B-\sec ^{2} A \tan ^{2} B=\frac{\sin ^{2} A}{\cos ^{2} A} \times \frac{1}{\cos ^{2} B}-\frac{1}{\cos ^{2} A} \times \frac{\sin ^{2} B}{\cos ^{2} B}$
$=\frac{\sin ^{2} A}{\cos ^{2} A \cos ^{2} B}-\frac{\sin ^{2} B}{\cos ^{2} A \cos ^{2} B}$
$=\frac{\sin ^{2} A-\sin ^{2} B}{\cos ^{2} A \cos ^{2} B}$
Let us consider RHS,
$\tan ^{2} A-\tan ^{2} B=\frac{\sin ^{2} A}{\cos ^{2} A}-\frac{\sin ^{2} B}{\cos ^{2} B}$
$=\frac{\sin ^{2} A \cos ^{2} B-\cos ^{2} A \sin ^{2} B}{\cos ^{2} A \cos ^{2} B}$
$=\frac{\sin ^{2} A\left(1-\sin ^{2} B\right)-\left(1-\sin ^{2} A\right) \sin ^{2} B}{\cos ^{2} A \cos ^{2} B}$
$=\frac{\sin ^{2} A-\sin ^{2} B}{\cos ^{2} A \cos ^{2} B}$
Here,
LHS $=$ RHS
Hence proved.
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