Trigonometrical Ratios and Identities 1 Question 4
4. If $\cos (\alpha+\beta)=\frac{3}{5}, \sin (\alpha-\beta)=\frac{5}{13}$ and $0<\alpha, \beta<\frac{\pi}{4}$, then $\tan (2 \alpha)$ is equal to
(2019 Main, 8 April I)
(a) $\frac{63}{52}$
(b) $\frac{63}{16}$
(c) $\frac{21}{16}$
(d) $\frac{33}{52}$
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Answer:
Correct Answer: 4. (c)
Solution:
- 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 $0<\beta<\frac{\pi}{4}$
$$ \begin{aligned} & \therefore \quad 0<\alpha+\beta<\frac{\pi}{4}+\frac{\pi}{4}=\frac{\pi}{2} \\ & \Rightarrow \quad 0<\alpha+\beta<\frac{\pi}{2} \end{aligned} $$
Also, $\quad-\frac{\pi}{4}<-\beta<0$
$\therefore \quad 0-\frac{\pi}{4}<\alpha-\beta<\frac{\pi}{4}+0$
$\Rightarrow \quad-\frac{\pi}{4}<\alpha-\beta<\frac{\pi}{4}$
$\therefore \alpha+\beta \in 0, \frac{\pi}{2}$ and $\alpha-\beta \in-\frac{\pi}{4}, \frac{\pi}{4}$
But $\sin (\alpha-\beta)>0$, therefore $\alpha-\beta \in 0, \frac{\pi}{4}$.
Now, $\quad \sin (\alpha-\beta)=\frac{5}{13}$
$\Rightarrow \quad \tan (\alpha-\beta)=\frac{5}{12}$
and $\quad \cos (\alpha+\beta)=\frac{3}{5}$
$\Rightarrow \quad \tan (\alpha+\beta)=\frac{4}{3}$
Now, $\tan (2 \alpha)=\tan [(\alpha+\beta)+(\alpha-\beta)]$
$=\frac{\tan (\alpha+\beta)+\tan (\alpha-\beta)}{1-\tan (\alpha+\beta) \tan (\alpha-\beta)}=\frac{\frac{4}{3}+\frac{5}{12}}{1-\frac{4}{3} \times \frac{5}{12}}$
[from Eqs. (i) and (ii)]
$=\frac{48+15}{36-20}=\frac{63}{16}$
