Question

If $\frac{\sin 3\theta }{\sin \theta }=\frac{11}{25}$ then $\tan \frac{\theta }{2}$ can have the value equal to

• A. $2$
• B. $\frac{1}{2}$
• C. $-2$
• D. $-\frac{1}{2}$

Answer 1

Answer: (A), (B), (C), (D)

The correct options are
A $2$
B $-2$
C $-\frac{1}{2}$
D $\frac{1}{2}$

$\frac{\sin 3x}{\sin x}=\frac{11}{25}$

$\frac{3\sin x-4{\sin }^{3}x}{\sin x}=\frac{11}{25}$

$3-4{\sin }^{2}x=\frac{11}{25}$

$4{\sin }^{2}x=\frac{64}{25}$

${\sin }^{2}x=\frac{16}{25}$

$\sin x=\pm \frac{4}{5}$
Hence
If
$\sin x=\frac{4}{5}$
Now
$\sin x=\frac{2\tan \frac{x}{2}}{1+{\tan }^2\frac{x}{2}}$
Let
$\tan \left(\frac{x}{2}\right)=t$
Hence
$4+4t^2=10t$
$2t^2-5t+2=0$
$t=2$ and $t=\frac{1}{2}$
Similarly if $\sin x=\frac{-4}{5}$
we get
$-2t^2-5t-2=0$
$2t^2+5t+2=0$
$t=\frac{-1}{2}$ and $t=-2$.