Prove the given trigonometric equation — task. Maths CBSE Live product, Class 10 (2026-27).

Prove that

\tan^{2} X - \tan^{2} Y

\(=\)

\frac{\sin^{2} X - \sin^{2} Y}{\cos^{2} X \cos^{2} Y}

Proof:

\begin{array}{l} LHS = \tan^{2} A - \tan^{2} B \\ = \frac{i^{2} A}{i^{2} A} - \frac{i^{2} B}{i^{2} B} \\ = \frac{i^{2} A i^{2} B - i^{2} B . i^{2} A}{i^{2} A . i^{2} B} \\ = \frac{i^{2} A (1 - i^{2} B) - \sin^{2} B (1 - i^{2} A)}{i^{2} A . i^{2} B} \\ = \frac{\sin^{2} A - \sin^{2} {A i}^{2} B - i^{2} B + \sin^{2} {B i}^{2} A}{i^{2} A . i^{2} B} \\ = \frac{i^{2} A - i^{2} B}{i^{2} A . i^{2} B} \end{array}

\(=RHS\)

Hence proved.

Did you find an error?

1. Prove the given trigonometric equation