# Prove the following identities:$\frac{\tan ^{3} \theta}{1+\tan ^{2} \theta}+\frac{\cot ^{3} \theta}{1+\cot ^{2} \theta}=\sec \theta \operatorname{cosec} \theta-2 \sin \theta \cos \theta$

To do:

We have to prove that $\frac{\tan ^{3} \theta}{1+\tan ^{2} \theta}+\frac{\cot ^{3} \theta}{1+\cot ^{2} \theta}=\sec \theta \operatorname{cosec} \theta-2 \sin \theta \cos \theta$.

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,

$\frac{\tan ^{3} \theta}{1+\tan ^{2} \theta}+\frac{\cot ^{3} \theta}{1+\cot ^{2} \theta}=\frac{\tan ^{3} \theta}{\sec ^{2} \theta}+\frac{\cot ^{3} \theta}{\operatorname {cosec} ^{2} \theta}$

$=\frac{\sin ^{3} \theta}{\cos ^{3} \theta}\times\frac{\cos ^{2} \theta}{1}+\frac{\cos ^{3} \theta}{\sin ^{3} \theta}\times\frac{\sin ^{2} \theta}{1}$
$=\frac{\sin ^{3} \theta}{\cos \theta}+\frac{\cos ^{3} \theta}{sin \theta}$
$=\frac{\sin ^{4} \theta+\cos ^{4} \theta}{\cos \theta \sin \theta}$

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

$=\frac{1-2 \sin ^{2} \theta \cos ^{2} \theta}{\cos \theta \sin \theta}$

$=\frac{1}{\cos \theta \sin \theta}-\frac{2 \sin ^{2} \theta \cos ^{2} \theta}{\cos \theta \sin \theta}$

$=\frac{1}{\cos \theta \sin \theta}-2 \sin \theta \cos \theta$

$=\sec \theta \operatorname{cosec} \theta-2 \sin \theta \cos \theta$

Hence proved.

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Updated on: 10-Oct-2022

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