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

PUMPA - SMART LEARNING

எங்கள் ஆசிரியர்களுடன் 1-ஆன்-1 ஆலோசனை நேரத்தைப் பெறுங்கள். டாப்பர் ஆவதற்கு நாங்கள் பயிற்சி அளிப்போம்

Book Free Demo

Answer variants:

\frac{\sin^{2} x \cos^{2} y - \sin^{2} y \cos^{2} x}{\cos^{2} x \cos^{2} y}

\frac{\sin^{2} x (1 - \sin^{2} y) - \sin^{2} y (1 - \sin^{2} x)}{\cos^{2} x \cos^{2} y}

\frac{\sin^{2} x + \sin^{2} x \sin^{2} y - \sin^{2} y + {\sin^{2} y \sin}^{2} x}{\cos^{2} x \cos^{2} y}

\frac{\sin^{2} x}{\cos^{2} x} - \frac{\sin^{2} y}{\cos^{2} y}

\frac{\sin^{2} x - \sin^{2} x \sin^{2} y - \sin^{2} y + {\sin^{2} y \sin}^{2} x}{\cos^{2} x \cos^{2} y}

\frac{\sin^{2} x \cos^{2} x - \sin^{2} y \cos^{2} y}{\cos^{2} x \cos^{2} y}

Show that

\tan^{2} x - \tan^{2} y

\(=\)

\frac{\sin^{2} x - \sin^{2} y}{\cos^{2} x \cos^{2} y}

Proof:

LHS \(=\)

\tan^{2} x - \tan^{2} y

\(=\)

\frac{\sin^{2} x - \sin^{2} y}{\cos^{2} x \cos^{2} y}

\(=\) RHS

Did you find an error?

8. Prove the given trigonometric equation