Question

Given that $\cos (\alpha -\beta )/2=2\cos (\alpha +\beta )/2,$ then $\tan \alpha /2\tan \beta /2$ is equal to

• A. $1/2$
• B. $1/3$
• C. $1/4$
• D. $1/8$

Answer 1

The correct option is B

$1/3$

Explanation for the correct option:

Find the value of $\tan \alpha /2\tan \beta /2$:

Given,

$$\begin{gathered} \Rightarrow \cos (\alpha –\beta )/2=2\cos (\alpha +\beta )/2 \\ \Rightarrow \frac{\left[\cos \right(\alpha –\beta )/2]}{\left[\cos \right(\alpha +\beta )/2]}=2 \\ \Rightarrow \frac{[\cos \alpha /2\cos \beta /2+\sin \alpha /2\sin \beta /2]}{[\cos \alpha /2\cos \beta /2–\sin \alpha /2\sin \beta /2]}=2 \end{gathered}$$

Dividing the numerator and denominator of LHS by $\cos \alpha /2\cos \beta /2$,

$$\begin{gathered} \Rightarrow \frac{[1+\tan \alpha /2\tan \beta /2]}{[1–\tan \alpha /2\tan \beta /2]}=2 \\ \Rightarrow 1+\tan \alpha /2\tan \beta /2=2–2\tan \alpha /2\tan \beta /2 \\ \Rightarrow \tan \alpha /2\tan \beta /2+2\tan \alpha /2\tan \beta /2=2–1 \\ \Rightarrow 3\tan \alpha /2\tan \beta /2=1 \\ \Rightarrow \tan \alpha /2\tan \beta /2=1/3 \end{gathered}$$

Hence the correct option is B.