Prove the following trigonometric identities.

In the given question, we need to prove $\frac{\tan ^{3} \theta}{1+\tan ^{2} \theta}+\frac{\cot ^{3} \theta}{1+\cot ^{2} \theta}=\sec \theta \operatorname{cosec} \theta-2 \sin \theta \cos \theta$

Using the property $1+\tan ^{2} \theta=\sec ^{2} \theta$ and $1+\cot ^{2} \theta=\operatorname{cosec}^{2} \theta$, we get

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

$=\frac{\left(\frac{\sin ^{3} \theta}{\cos ^{3} \theta}\right)}{\left(\frac{1}{\cos ^{2} \theta}\right)}+\frac{\left(\frac{\cos ^{3} \theta}{\sin ^{3} \theta}\right)}{\left(\frac{1}{\sin ^{2} \theta}\right)}$

Taking the reciprocal of the denominator, we get

$\frac{\left(\frac{\sin ^{3} \theta}{\cos ^{3} \theta}\right)}{\left(\frac{1}{\cos ^{2} \theta}\right)}+\frac{\left(\frac{\cos ^{3} \theta}{\sin ^{3} \theta}\right)}{\left(\frac{1}{\sin ^{2} \theta}\right)}=\left(\frac{\sin ^{3} \theta}{\cos ^{3} \theta} \times \frac{\cos ^{2} \theta}{1}\right)+\left(\frac{\cos ^{3} \theta}{\sin ^{3} \theta} \times \frac{\sin ^{2} \theta}{1}\right)$

$=\frac{\sin ^{3} \theta}{\cos \theta}+\frac{\cos ^{3} \theta}{\sin \theta}$

$=\frac{\sin ^{4} \theta+\cos ^{4} \theta}{\cos \theta \sin \theta}$

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

Further, using the identity $a^{2}+b^{2}=(a+b)^{2}-2 a b$, we get

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

$=\frac{1}{\cos \theta \sin \theta}-\frac{2 \sin ^{2} \theta \cos ^{2} \theta}{\cos \theta \sin \theta}\left(\right.$ using $\left.\sin ^{2} \theta+\cos ^{2} \theta=1\right)$

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

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

Hence proved.