Aspire Faculty ID #3649 · Topic: NIMCET 2012 · Just now
NIMCET 2012

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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