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