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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.
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