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SSC CHSL 10 Jan 2017 Shift 2 Paper

Question : 81 of 100

Marks: +1, -0

What is the value of tan3A?

Solution:

Expression :

\tan(3A)

\frac{\sin(3A)}{\cos(3A)} = \frac{\sin(2A+A)}{\cos(2A+A)}

\frac{\sin(2A)\cos A + \cos(2A)\sin A}{\cos(2A)\cos A - \sin(2A)\sin A}

Using,

\sin(2x) = 2\sin x \cos x

and

\cos(2x) = \cos^2 x - \sin^2 x

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

Taking common from numerator and from denominator,

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

\tan A \, \frac{3\cos^2 A - \sin^2 A}{\cos^2 A - 3\sin^2 A}

Dividing both numerator and denominator by

\cos^2 A

, we get :

\tan A \, \frac{3 - \tan^2 A}{1 - 3\tan^2 A}

\frac{3\tan A - \tan^3 A}{1 - 3\tan^2 A}

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