Prove the given trigonometric equation — task. Maths CBSE Homework Assignments, Class 10.

Answer variants:

\frac{\sin^{2} A - \sin^{2} A \sin^{2} B - \sin^{2} B + {\sin^{2} B \sin}^{2} A}{\cos^{2} A \cos^{2} B}

\frac{\sin^{2} A + \sin^{2} A \sin^{2} B - \sin^{2} B + {\sin^{2} B \sin}^{2} A}{\cos^{2} A \cos^{2} B}

\frac{\sin^{2} A \cos^{2} A - \sin^{2} B \cos^{2} B}{\cos^{2} A \cos^{2} B}

\frac{\sin^{2} A (1 - \sin^{2} B) - \sin^{2} B (1 - \sin^{2} A)}{\cos^{2} A \cos^{2} B}

\frac{\sin^{2} A}{\cos^{2} A} - \frac{\sin^{2} B}{\cos^{2} B}

\frac{\sin^{2} A \cos^{2} B - \sin^{2} B \cos^{2} A}{\cos^{2} A \cos^{2} B}

Prove that

\tan^{2} A - \tan^{2} B

\(=\)

\frac{\sin^{2} A - \sin^{2} B}{\cos^{2} A \cos^{2} B}

Proof:

LHS \(=\)

\tan^{2} A - \tan^{2} B

\(=\)

\frac{\sin^{2} A - \sin^{2} B}{\cos^{2} A \cos^{2} B}

\(=\) RHS

Did you find an error?

1. Prove the given trigonometric equation