TBQ - Prove the given equation — task. Maths TNSB Mentoring, Class 10 Crash course.

Establish that \(tan^2 \ A - tan^2 \ B = \frac{sin^2 \ A - sin^2 \ B}{cos^2 \ A \ cos^2 \ B}\).

Proof:

Answer variants:

\(1 - sin^2 A\)

\(sin^2 A - sin^2 B\)

\(1 - sin^2 B\)

\(1 - cosec^2 A\)

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

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9. TBQ - Prove the given equation