Question

if $\cos (\alpha +\beta )=\frac{4}{5};\sin (\alpha -\beta )=\frac{5}{13}$ and $\alpha ,\beta$ lie between $0$ and $\frac{\pi }{4}$, then find the value of $\tan 2\alpha$

• A. $\frac{56}{33}$
• B. $\frac{26}{33}$
• C. $\frac{36}{33}$
• D. None of these

Answer 1

The correct option is A $\frac{56}{33}$

Given

$\cos (\alpha +\beta )=\frac{4}{5}⟹\sin (\alpha +\beta )=\frac{3}{5}$

$⟹\tan (\alpha +\beta )=\frac{3}{4}\cdots \left(1\right)$

$\sin (\alpha -\beta )=\frac{5}{13}⟹\cos (\alpha -\beta )=\frac{12}{13}$

$⟹\tan (\alpha -\beta )=\frac{5}{12}\cdots \left(2\right)$

Consider

$\tan \left(2\alpha \right)$

$\tan (\alpha +\beta +\alpha -\beta )$

$\frac{\tan (\alpha +\beta )+\tan (\alpha -\beta )}{1-\tan (\alpha +\beta )\tan (\alpha -\beta )}$

From $\left(1\right)\&\left(2\right)$

$⟹\frac{\frac{3}{4}+\frac{5}{12}}{1-(\frac{3}{4}\times \frac{5}{12})}$

$⟹\frac{\frac{36+20}{48}}{\frac{48-15}{48}}$

$⟹\frac{56}{33}$