Question

If $\alpha ,\beta$ are the different values of $3\cos \theta +4\sin \theta =\frac{9}{2}$ and $A=\tan (\frac{\alpha }{2}+\frac{\beta }{2}),B=\tan \frac{\alpha }{2}\tan \frac{\beta }{2},C=\sin (\alpha +\beta ),$ then which one of the option is true?

• A. $A>B>C$
• B. $C>B>A$
• C. $A>C>B$
• D. $A=B=C$

Answer 1

The correct option is C $A>C>B$

$3\cos \theta +4\sin \theta =\frac{9}{2}$

$\Rightarrow 6-6{\tan }^2\frac{\theta }{2}+16\tan \frac{\theta }{2}=9+9{\tan }^2\frac{\theta }{2}$

$\Rightarrow 15{\tan }^2\frac{\theta }{2}-16\tan \frac{\theta }{2}+3=0$

This is quadratic in $\tan x$ and $\tan \alpha ,\tan \beta$ are its roots.
$\Rightarrow \tan \frac{\alpha }{2}+\tan \frac{\beta }{2}=\frac{16}{15}$ and $\tan \frac{\alpha }{2}\tan \frac{\beta }{2}=\frac{3}{15}$

Using these, A) $\tan \frac{\alpha +\beta }{2}=\frac{\tan \frac{\alpha }{2}+\tan \frac{\beta }{2}}{1-\tan \frac{\alpha }{2}\tan \frac{\beta }{2}}=\frac{4}{3}$

B) $\tan \frac{\alpha }{2}\tan \frac{\beta }{2}=\frac{3}{15}$

C) $\sin (\alpha +\beta )=\frac{2\tan \frac{\alpha +\beta }{2}}{1+{\tan }^2\frac{\alpha +\beta }{2}}=\frac{24}{25}$

Hence $A>C>B$.