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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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