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Study Stuff for MCA Examinations - NIMCET
Study Stuff for MCA Examinations - NIMCET
Previous Year Question (PYQs)
If $ \cos(\alpha+\beta)=\frac{4}{5}$ and $\sin(\alpha-\beta)=\frac{5}{13},\quad 0<\alpha,\beta<\frac{\pi}{4},$ then $ \tan(2\alpha)$ is equal to
Solution
Given:$\cos(\alpha+\beta)=\frac{4}{5}$
Since $0<\alpha,\beta<\frac{\pi}{4}$,
therefore $\alpha+\beta$ is an acute angle.
$\sin(\alpha+\beta)=\sqrt{1-\cos^2(\alpha+\beta)}$
$=\sqrt{1-\frac{16}{25}}=\frac{3}{5}$
Therefore,
$\tan(\alpha+\beta)=\frac{3}{4}$
Also,
$\sin(\alpha-\beta)=\frac{5}{13}$
Since the sine value is positive, $\alpha-\beta>0$.
Therefore,
$\cos(\alpha-\beta)=\sqrt{1-\sin^2(\alpha-\beta)}$
$=\sqrt{1-\frac{25}{169}}=\frac{12}{13}$
Hence,
$\tan(\alpha-\beta)=\frac{5}{12}$
Now,
$2\alpha=(\alpha+\beta)+(\alpha-\beta)$
Using the tangent addition formula:
$\tan(2\alpha)=\frac{\tan(\alpha+\beta)+\tan(\alpha-\beta)}{1-\tan(\alpha+\beta)\tan(\alpha-\beta)}$
$=\frac{\frac{3}{4}+\frac{5}{12}}{1-\left(\frac{3}{4}\right)\left(\frac{5}{12}\right)}$
$=\frac{\frac{14}{12}}{1-\frac{15}{48}}$
$=\frac{\frac{7}{6}}{\frac{33}{48}}$
$=\frac{7}{6}\times\frac{48}{33}$
$=\frac{56}{33}$
Therefore, the correct answer is:
$\frac{56}{33}$
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