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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Updated on: 10-Oct-2022

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