Question

If $\tan\alpha = \frac{1}{7}$and $\sin\beta = \frac{1}{\sqrt{10}}\left( 0 < \alpha,\beta < \frac{\pi}{2} \right)$, then $2\beta$is equal to

• A. $\frac{\pi}{4} - \alpha$
• B. $\frac{3\pi}{4} - \alpha$
• C. $\frac{\pi}{8} - \frac{\alpha}{2}$
• D. $\frac{3\pi}{8} - \frac{\alpha}{2}$

Answer 1

Answer: (A)

$\frac{\pi}{4} - \alpha$

Answer 2

Since $\sin\beta = \frac{1}{\sqrt{10}} \Rightarrow \tan\beta = \frac{1}{3}$

⇒ $\tan 2\beta = \frac{2\tan\beta}{1 - \tan^{2}\beta} = \frac{3}{4}$

$\therefore\tan(\alpha + 2\beta) = \frac{\frac{1}{7} + \frac{3}{4}}{1 - \frac{1}{7}.\frac{3}{4}} = \frac{25}{25} = 1$

Now,$0 < \beta < \frac{\pi}{2}$and$\tan 2\beta = \frac{3}{4} > 0$both⇒ $0 < 2\beta < \frac{\pi}{2}$.

Again,$0 < \alpha < \frac{\pi}{2}$and$0 < 2\beta < \frac{\pi}{2}$both⇒ $0 < \alpha + 2\beta < \pi$

Thus, $0 < \alpha + 2\beta < \pi$and $\tan(\alpha + 2\beta) = 1$both

⇒ $\alpha + 2\beta = \frac{\pi}{4} \Rightarrow 2\beta = \frac{\pi}{4} - \alpha$.