Prove that: $\left(1 + \frac{1}{ an^2 heta} ight)\left(1 + \frac{1}{\cot^2 heta} ight) = \frac{1}{\sin^2 heta - \sin^4 heta}$

Question

Prove that: $\left(1 + \frac{1}{	an^2	heta}ight)\left(1 + \frac{1}{\cot^2	heta}ight) = \frac{1}{\sin^2	heta - \sin^4	heta}$

Accepted Answer

Step 1: LHS $= (1 + \cot^2	heta)(1 + 	an^2	heta) = \csc^2	heta \cdot \sec^2	heta$
Step 2: $= \frac{1}{\sin^2	heta} \cdot \frac{1}{\cos^2	heta} = \frac{1}{\sin^2	heta \cdot \cos^2	heta}$
Step 3: $= \frac{1}{\sin^2	heta(1 - \sin^2	heta)} = \frac{1}{\sin^2	heta - \sin^4	heta}$ = RHS

Question

Solution