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Prove the following trigonometric identities:$ \frac{\left(1+\cot ^{2} \theta\right) \tan \theta}{\sec ^{2} \theta}=\cot \theta $
To do:
We have to prove that \( \frac{\left(1+\cot ^{2} \theta\right) \tan \theta}{\sec ^{2} \theta}=\cot \theta \).
Solution:
We know that,
$\operatorname{cosec}^2 \theta-cot ^{2} \theta=1$.......(i)
$\cot \theta=\frac{\operatorname{cosec} \theta}{\sec \theta}$........(ii)
$\tan \theta \times \cot \theta=1$........(iii)
Therefore,
$\frac{\left(1+\cot ^{2} \theta\right) \tan \theta}{\sec ^{2} \theta}=\frac{\operatorname{cosec} ^{2} \theta) \tan \theta}{\sec ^{2} \theta}$ (From (i))
$=(\frac{\operatorname{cosec} \theta}{\sec \theta})^2\times \tan \theta$
$=(\cot \theta)^2\times \tan \theta$ (From (ii))
$=\cot \theta \times \cot \theta \times \tan \theta$
$=1\times \cot \theta$ (From (iii))
$=\cot \theta$
Hence proved.