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Prove that:$ \tan ^{2} A+\cot ^{2} A=\sec ^{2} A \operatorname{cosec}^{2} A-2 $
To do:
We have to prove that \( \tan ^{2} A+\cot ^{2} A=\sec ^{2} A \operatorname{cosec}^{2} A-2 \).
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,
$\tan ^{2} \mathrm{~A}+\cot ^{2} \mathrm{~A}=\frac{\sin ^{2} \mathrm{~A}}{\cos ^{2} \mathrm{~A}}+\frac{\cos ^{2} \mathrm{~A}}{\sin ^{2} \mathrm{~A}}$
$=\frac{\sin ^{4} \mathrm{~A}+\cos ^{4} \mathrm{~A}}{\sin ^{2} \mathrm{~A} \cos ^{2} \mathrm{~A}}$
$=\frac{\left(\sin ^{2} \mathrm{~A}+\cos ^{2} \mathrm{~A}\right)^{2}-2 \sin ^{2} \mathrm{~A} \cos ^{2} \mathrm{~A}}{\sin ^{2} \mathrm{~A} \cos ^{2} \mathrm{~A}}$
$=\frac{(1)^{2}-2 \sin ^{2} A \cos ^{2} A}{\sin ^{2} A \cos ^{2} A}$
$=\frac{1-2 \sin ^{2} A \cos ^{2} A}{\sin ^{2} A \cos ^{2} A}$
$=\frac{1}{\sin ^{2} A \cos ^{2} A}-\frac{2 \sin ^{2} A \cos ^{2} A}{\sin ^{2} A \cos ^{2} A}$
$=\sec ^{2} A \operatorname{cosec}^{2} A-2$
Hence proved.
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