Question

If $\cos\alpha = \frac{l\cos\beta+m}{l+m\cos\beta}$, then $\frac{\tan^2(\alpha/2)}{\tan^2(\beta/2)} =$

• A. $\frac{l-m}{l+m}$
• B. $\frac{l+m}{l-m}$
• C. $\frac{l^2-m^2}{l^2+m^2}$
• D. $\frac{l-m}{l+m}$ (Duplicate option)

Hint:

The half-angle identity $\tan^2(\theta/2) = \frac{1-\cos\theta}{1+\cos\theta}$ is extremely useful for problems that relate trigonometric functions of an angle $\alpha$ to functions of an angle $\beta$ through a formula for $\cos\alpha$.

Answer 1

The Correct Option is A

Step 1: Use half-angle identity \[ \tan^2(\theta/2) = \frac{1-\cos\theta}{1+\cos\theta} \]
Step 2: Substitute $\cos\alpha = \frac{l\cos\beta + m{l + m \cos\beta}$.}
\[ \tan^2(\alpha/2) = \frac{1-\cos\alpha}{1+\cos\alpha} = \frac{(l-m)(1-\cos\beta)}{(l+m)(1+\cos\beta)} \]
Step 3: Express in terms of $\tan^2(\beta/2)$.
\[ \tan^2(\beta/2) = \frac{1-\cos\beta}{1+\cos\beta} \implies \frac{\tan^2(\alpha/2)}{\tan^2(\beta/2)} = \frac{l-m}{l+m} \]