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.