Question

If ${\theta }_1$ ad ${\theta }_2$ are two values lying in $[0,2\pi ]$ for which $\tan \theta =\lambda$, then $\tan \frac{{\theta }_1}{2}.\tan \frac{{\theta }_2}{2}$ is

• A. Zero
• B. $-1$
• C. $2$
• D. $1$

Answer 1

The correct option is B $-1$

Given $\tan \theta =\lambda$

$\frac{2\tan \frac{\theta }{2}}{1-{\tan }^2\frac{\theta }{2}}=\lambda$

$⟹1-{\tan }^2\frac{\theta }{2}-\frac{2}{\lambda }\tan \frac{\theta }{2}=0$

$⟹{\tan }^2\frac{\theta }{2}+\frac{2}{\lambda }\tan \frac{\theta }{2}-1=0$

Given ${\theta }_1,{\theta }_2$ are the two solutions for this equation

So $\tan \frac{{\theta }_1}{2}\tan \frac{{\theta }_2}{2}=\frac{-1}{1}=-1$