Prove the following trigonometric identities.

Solution:

In the given question, we need to prove

$\left(\frac{1}{\sec ^{2} \theta-\cos ^{2} \theta}+\frac{1}{\operatorname{cosec}^{2} \theta-\sin ^{2} \theta}\right) \sin ^{2} \theta \cos ^{2} \theta=\left(\frac{1-\sin ^{2} \theta \cos ^{2} \theta}{2+\sin ^{2} \theta \cos ^{2} \theta}\right)$

Now, using $\sec \theta=\frac{1}{\cos \theta}$ and $\operatorname{cosec} \theta=\frac{1}{\sin \theta}$ in L.H.S, we get

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

$=\left(\frac{1}{\left(\frac{1-\cos ^{4} \theta}{\cos ^{2} \theta}\right)}+\frac{1}{\left(\frac{1-\sin ^{4} \theta}{\sin ^{2} \theta}\right)}\right) \sin ^{2} \theta \cos ^{2} \theta$

$=\left(\frac{\cos ^{2} \theta}{1-\cos ^{4} \theta}+\frac{\sin ^{2} \theta}{1-\sin ^{4} \theta}\right) \sin ^{2} \theta \cos ^{2} \theta$

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

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

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

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

$=\left(\frac{\cos ^{4} \theta\left(1+\sin ^{2} \theta\right)+\sin ^{4} \theta\left(1+\cos ^{2} \theta\right)}{\left(1+\cos ^{2} \theta\right)\left(1+\sin ^{2} \theta\right)}\right)$

Further using the identity $\sin ^{2} \theta+\cos ^{2} \theta=1$, we get

L.H.S. $=\left(\frac{\cos ^{4} \theta+\cos ^{4} \theta \sin ^{2} \theta+\sin ^{4} \theta+\sin ^{4} \theta \cos ^{2} \theta}{1+\cos ^{2} \theta+\sin ^{2} \theta+\sin ^{2} \theta \cos ^{2} \theta}\right)$

$=\left(\frac{\cos ^{4} \theta+\sin ^{4} \theta+\cos ^{2} \theta \sin ^{2} \theta\left(\cos ^{2} \theta+\sin ^{2} \theta\right)}{2+\sin ^{2} \theta \cos ^{2} \theta}\right)$

$=\left(\frac{\cos ^{4} \theta+\sin ^{4} \theta+\cos ^{2} \theta \sin ^{2} \theta(1)}{2+\sin ^{2} \theta \cos ^{2} \theta}\right)$

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

So,

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

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

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

Hence proved.