Prove the following trigonometric identities:$ \frac{1+\tan ^{2} \theta}{1+\cot ^{2} \theta}=\left(\frac{1-\tan \theta}{1-\cot \theta}\right)^{2}=\tan ^{2} \theta $

To do:

We have to prove that \( \frac{1+\tan ^{2} \theta}{1+\cot ^{2} \theta}=\left(\frac{1-\tan \theta}{1-\cot \theta}\right)^{2}=\tan ^{2} \theta \).

Solution:

We know that,

$\cos ^{2} \theta+\sin^2 \theta=1$.......(i)

$\sec^2 \theta-\tan^2 \theta=1$.......(ii)

$\operatorname{cosec} ^2 \theta-\cot^2 \theta=1$......(iii)

$\tan \theta=\frac{\sin \theta}{\cos \theta}$.........(iv)

$\cot \theta=\frac{\cos \theta}{\sin \theta}$.........(v)

Therefore,

$\frac{1+\tan ^{2} \theta}{1+\cot ^{2} \theta}=\frac{1+\frac{\sin^2 \theta}{\cos^2 \theta}}{1+\frac{\cos^2 \theta}{\sin^2 \theta}}$

$=\frac{\frac{\sin^2 \theta+\cos^2 \theta}{\cos^2 \theta}}{\frac{\cos^2 \theta+\sin^2 \theta}{\sin^2 \theta}}$   

$=\frac{\sin^2 \theta}{\cos^2 \theta}$               

$=\tan^2 \theta$          

$(\frac{1-\tan \theta}{1-\cot \theta})^{2}=(\frac{1-\frac{\sin \theta}{\cos \theta}}{1-\frac{\cos \theta}{\sin \theta}})^{2}$         

$=(\frac{\frac{\cos \theta-\sin \theta}{\cos \theta}}{\frac{\sin \theta-\cos \theta}{\sin \theta}})^{2}$

$=(\frac{-\sin \theta}{\cos \theta})^2$

$=(-\tan \theta)^2$

$=\tan^2 \theta$

Hence proved.