Prove the given trigonometric equation — task. Mathematics CBSE, Class 10.

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Answer variants:

\frac{\sin^{2} α (1 - \sin^{2} β) - \sin^{2} β (1 - \sin^{2} α)}{\cos^{2} α \cos^{2} β}

\frac{\sin^{2} α}{\cos^{2} α} - \frac{\sin^{2} β}{\cos^{2} β}

\frac{\sin^{2} α \cos^{2} α - \sin^{2} β \cos^{2} β}{\cos^{2} α \cos^{2} β}

\frac{\sin^{2} α + \sin^{2} α \sin^{2} β - \sin^{2} β + {\sin^{2} β \sin}^{2} α}{\cos^{2} α \cos^{2} β}

\frac{\sin^{2} α \cos^{2} β - \sin^{2} β \cos^{2} α}{\cos^{2} α \cos^{2} β}

\frac{\sin^{2} α - \sin^{2} α \sin^{2} β - \sin^{2} β + {\sin^{2} β \sin}^{2} α}{\cos^{2} α \cos^{2} β}

Show that

\tan^{2} α - \tan^{2} β

\(=\)

\frac{\sin^{2} α - \sin^{2} β}{\cos^{2} α \cos^{2} β}

Proof:

LHS \(=\)

\tan^{2} α - \tan^{2} β

\(=\)

\frac{\sin^{2} α - \sin^{2} β}{\cos^{2} α \cos^{2} β}

\(=\) RHS

Did you find an error?

7. Prove the given trigonometric equation