Question

Prove that \(\left(1 + \frac{1}{\tan^2 \theta}\right) \left(1 + \frac{1}{\cot^2 \theta}\right) = \frac{1}{\sin^2 \theta - \sin^4 \theta}\).

Answer 1

Trigonometric Identity Proof 

We need to prove the identity:

\[ \left(1 + \frac{1}{\tan^2 \theta}\right) \left(1 + \frac{1}{\cot^2 \theta}\right) = \frac{1}{\sin^2 \theta - \sin^4 \theta} \]

Left Hand Side (LHS):

\[ \left(1 + \frac{1}{\tan^2 \theta}\right) \left(1 + \frac{1}{\cot^2 \theta}\right) \] Use identities: \[ \frac{1}{\tan^2 \theta} = \cot^2 \theta, \quad \frac{1}{\cot^2 \theta} = \tan^2 \theta \] So: \[ \left(1 + \cot^2 \theta\right)\left(1 + \tan^2 \theta\right) \] Using Pythagorean identities: \[ 1 + \cot^2 \theta = \csc^2 \theta, \quad 1 + \tan^2 \theta = \sec^2 \theta \] Therefore: \[ \text{LHS} = \csc^2 \theta \cdot \sec^2 \theta = \frac{1}{\sin^2 \theta} \cdot \frac{1}{\cos^2 \theta} = \frac{1}{\sin^2 \theta \cos^2 \theta} \]

Right Hand Side (RHS):

\[ \frac{1}{\sin^2 \theta - \sin^4 \theta} \] Factor denominator: \[ = \frac{1}{\sin^2 \theta (1 - \sin^2 \theta)} = \frac{1}{\sin^2 \theta \cos^2 \theta} \] (Using identity \(1 - \sin^2 \theta = \cos^2 \theta\))

Conclusion:

\[ \text{LHS} = \text{RHS} = \frac{1}{\sin^2 \theta \cos^2 \theta} \] \[ \boxed{\text{LHS = RHS, Proved}} \]