Question

If $\cos (\alpha +\beta )=45,$ $\sin (\alpha -\beta )=513$ and $\alpha ,\beta$ lie between $0$ and $\frac{\pi }{4},$ find $\tan 2\alpha .$

• A. $\frac{63}{16}$
• B. $\frac{21}{16}$
• C. $\frac{63}{52}$
• D. $\frac{33}{52}$

Answer 1

The correct option is A $\frac{63}{16}$

Given, $\sin (\alpha -\beta )=\frac{5}{13}$ and $\cos (\alpha +\beta )=\frac{3}{5}$, where $\alpha ,\beta \in (0,\frac{\pi }{4})$ Since, $0<\alpha <\frac{\pi }{4}$ and $0<\beta <\frac{\pi }{4}$
$\therefore 0<\alpha +\beta <\frac{\pi }{4}+\frac{\pi }{4}=\frac{\pi }{2}$
$\Rightarrow 0<\alpha +\beta <\frac{\pi }{2}$
$\text{Also,}-\frac{\pi }{4}<-\beta <0$
$\therefore 0-\frac{\pi }{4}<\alpha -\beta <\frac{\pi }{4}+0$
$\therefore \alpha +\beta \in (0,\frac{\pi }{2})\text{and}\alpha -\beta \in (-\frac{\pi }{4},\frac{\pi }{4})$
$\text{But}\sin (\alpha -\beta )>0,\text{therefore}\alpha -\beta \in (0,\frac{\pi }{4})\text{.}$
Now, $\sin (\alpha -\beta )=\frac{5}{13}$
$\Rightarrow \tan (\alpha -\beta )=\frac{5}{12}...\left(i\right)$
and $\cos (\alpha +\beta )=\frac{3}{5}$
$\Rightarrow \tan (\alpha +\beta )=\frac{4}{3}...\left(ii\right)$
$\text{Now,}\tan \left(2\alpha \right)=\tan \left[\right(\alpha +\beta )+(\alpha -\beta \left)\right]$
$=\frac{\tan (\alpha +\beta )+\tan (\alpha -\beta )}{1-\tan (\alpha +\beta )\tan (\alpha -\beta )}=\frac{\frac{4}{3}+\frac{5}{12}}{1-\frac{4}{3}\times \frac{5}{12}}$
$\text{[from Eqs. (i) and (ii)]}$
$=\frac{48+15}{36-20}=\frac{63}{16}$