Q84 / 180 · COMEDK 2025 (10 May Shift 3) · Mathematics
If $\tan \alpha=\dfrac{1}{7}$ and $\sin \beta=\dfrac{1}{\sqrt{10}}, \quad 0 <\alpha, \beta <\dfrac{\pi}{2}$ then $2 \beta$ is equal to
- A$\dfrac{3 \pi}{4}-\alpha$
- B$\dfrac{3 \pi}{8}-\dfrac{\alpha}{2}$
- Correct answerC$\dfrac{\pi}{4}-\alpha$
- D$\dfrac{\pi}{8}-\alpha$
Solution
Given $\tan \alpha = \dfrac{1}{7}$ and $\sin \beta = \dfrac{1}{\sqrt{10}}$.
Since $\sin \beta = \dfrac{1}{\sqrt{10}}$, we have $\cos \beta = \sqrt{1 - \sin^2 \beta} = \sqrt{1 - \dfrac{1}{10}} = \sqrt{\dfrac{9}{10}} = \dfrac{3}{\sqrt{10}}$.
Therefore, $\tan \beta = \dfrac{\sin \beta}{\cos \beta} = \dfrac{1/\sqrt{10}}{3/\sqrt{10}} = \dfrac{1}{3}$.
We calculate $\tan 2\beta$ using the formula $\tan 2\beta = \dfrac{2 \tan \beta}{1 - \tan^2 \beta}$.
$\tan 2\beta = \dfrac{2(1/3)}{1 - (1/3)^2} = \dfrac{2/3}{1 - 1/9} = \dfrac{2/3}{8/9} = \dfrac{2}{3} \times \dfrac{9}{8} = \dfrac{3}{4}$.
Now consider $\tan(\alpha + 2\beta) = \dfrac{\tan \alpha + \tan 2\beta}{1 - \tan \alpha \tan 2\beta}$.
Substituting the values, $\tan(\alpha + 2\beta) = \dfrac{1/7 + 3/4}{1 - (1/7)(3/4)} = \dfrac{(4+21)/28}{(28-3)/28} = \dfrac{25/28}{25/28} = 1$.
Since $0 < \alpha, \beta < \dfrac{\pi}{2}$, we have $0 < \alpha + 2\beta < \dfrac{3\pi}{2}$. Given $\tan(\alpha + 2\beta) = 1$, the possible values for $\alpha + 2\beta$ are $\dfrac{\pi}{4}$ or $\dfrac{5\pi}{4}$.
Since $\tan \alpha = 1/7$ and $\tan \beta = 1/3$, both $\alpha$ and $\beta$ are small positive angles, so $\alpha + 2\beta$ must be $\dfrac{\pi}{4}$.
Thus, $2\beta = \dfrac{\pi}{4} - \alpha$.
Answer: $\dfrac{\pi}{4}-\alpha$
Since $\sin \beta = \dfrac{1}{\sqrt{10}}$, we have $\cos \beta = \sqrt{1 - \sin^2 \beta} = \sqrt{1 - \dfrac{1}{10}} = \sqrt{\dfrac{9}{10}} = \dfrac{3}{\sqrt{10}}$.
Therefore, $\tan \beta = \dfrac{\sin \beta}{\cos \beta} = \dfrac{1/\sqrt{10}}{3/\sqrt{10}} = \dfrac{1}{3}$.
We calculate $\tan 2\beta$ using the formula $\tan 2\beta = \dfrac{2 \tan \beta}{1 - \tan^2 \beta}$.
$\tan 2\beta = \dfrac{2(1/3)}{1 - (1/3)^2} = \dfrac{2/3}{1 - 1/9} = \dfrac{2/3}{8/9} = \dfrac{2}{3} \times \dfrac{9}{8} = \dfrac{3}{4}$.
Now consider $\tan(\alpha + 2\beta) = \dfrac{\tan \alpha + \tan 2\beta}{1 - \tan \alpha \tan 2\beta}$.
Substituting the values, $\tan(\alpha + 2\beta) = \dfrac{1/7 + 3/4}{1 - (1/7)(3/4)} = \dfrac{(4+21)/28}{(28-3)/28} = \dfrac{25/28}{25/28} = 1$.
Since $0 < \alpha, \beta < \dfrac{\pi}{2}$, we have $0 < \alpha + 2\beta < \dfrac{3\pi}{2}$. Given $\tan(\alpha + 2\beta) = 1$, the possible values for $\alpha + 2\beta$ are $\dfrac{\pi}{4}$ or $\dfrac{5\pi}{4}$.
Since $\tan \alpha = 1/7$ and $\tan \beta = 1/3$, both $\alpha$ and $\beta$ are small positive angles, so $\alpha + 2\beta$ must be $\dfrac{\pi}{4}$.
Thus, $2\beta = \dfrac{\pi}{4} - \alpha$.
Answer: $\dfrac{\pi}{4}-\alpha$
Hence, the correct answer is option: 'C'.