Question

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

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

Hint:

Convert given trig values into \( \tan \) formThen compare using identities like \( \tan(A-B) \).

Answer 1

The Correct Option is B

Step 1: Find \( \cos\beta \).
\[ \sin\beta=\frac{1}{\sqrt{10}} \Rightarrow \cos\beta=\sqrt{1-\frac{1}{10}}=\frac{3}{\sqrt{10}} \]

Step 2: Find \( \tan\beta \).
\[ \tan\beta=\frac{\sin\beta}{\cos\beta}=\frac{1}{3} \]

Step 3: Use double angle formula.
\[ \tan 2\beta=\frac{2\tan\beta}{1-\tan^2\beta} \]
\[ =\frac{2\cdot \frac{1}{3}}{1-\frac{1}{9}}=\frac{\frac{2}{3}}{\frac{8}{9}}=\frac{3}{4} \]

Step 4: Find \( \tan\alpha \).
Given:
\[ \tan\alpha=\frac{1}{7} \]

Step 5: Use identity.
\[ \tan\left(\frac{\pi}{4}-\alpha\right)=\frac{1-\tan\alpha}{1+\tan\alpha} \]
\[ =\frac{1-\frac{1}{7}}{1+\frac{1}{7}}=\frac{\frac{6}{7}}{\frac{8}{7}}=\frac{3}{4} \]

Step 6: Compare both values.
\[ \tan 2\beta = \tan\left(\frac{\pi}{4}-\alpha\right) \]

Step 7: Final conclusion.
Since angles are in first quadrant:
\[ \boxed{2\beta = \frac{\pi}{4}-\alpha} \]