Question

If $\tan(\frac{\pi}{4}+\frac{\alpha}{2}) = \tan^3(\frac{\pi}{4}+\frac{\beta}{2})$, then $\frac{3+\sin^2\beta}{1+3\sin^2\beta}=$

• A. $\frac{\cos\beta}{\cos\alpha}$
• B. $\frac{\cos^3\alpha}{\sin^3\beta}$
• C. $\frac{\sin\alpha}{\sin\beta}$
• D. $\frac{\cos\alpha}{\cos\beta}$

Hint:

The identity $\tan^2(\frac{\pi}{4}+\frac{\theta}{2}) = \frac{1+\sin\theta}{1-\sin\theta}$ is a very useful tool for converting between tangent of a shifted half-angle and the sine of the full angle. It simplifies many trigonometric problems involving such expressions.

Answer 1

The Correct Option is C

Step 1: Use the identity \[ \tan^2\Big(\frac{\pi}{4}+\frac{\theta}{2}\Big) = \frac{1+\sin\theta}{1-\sin\theta} \implies \sin\theta = \frac{\tan^2(\frac{\pi}{4}+\frac{\theta}{2})-1}{\tan^2(\frac{\pi}{4}+\frac{\theta}{2})+1}. \]
Step 2: Let $X=\tan(\frac{\pi}{4}+\frac{\alpha}{2})$, $Y=\tan(\frac{\pi}{4}+\frac{\beta}{2})$, so $X=Y^3$.
Step 3: Express sines.
\[ \sin\alpha = \frac{X^2-1}{X^2+1} = \frac{Y^6-1}{Y^6+1}, \sin\beta = \frac{Y^2-1}{Y^2+1} \]
Step 4: Compute the ratio.
\[ \frac{\sin\alpha}{\sin\beta} = \frac{Y^6-1}{Y^6+1} \cdot \frac{Y^2+1}{Y^2-1} = \frac{Y^4+Y^2+1}{Y^4-Y^2+1} \]
Step 5: Compute target expression.
\[ \frac{3+\sin^2\beta}{1+3\sin^2\beta} = \frac{4Y^4+4Y^2+4}{4Y^4-4Y^2+4} = \frac{Y^4+Y^2+1}{Y^4-Y^2+1} \]
Step 6: Conclusion.
\[ \frac{3+\sin^2\beta}{1+3\sin^2\beta} = \frac{\sin\alpha}{\sin\beta} \]