Question

If $\tan x=-\frac{4}{3},\frac{\pi }{2}<x<\pi ,$ then which of the following is/are correct ?

• A. $\sin \frac{x}{2}=\frac{2\sqrt{5}}{5}$
• B. $\cos \frac{x}{2}=\frac{\sqrt{5}}{5}$
• C. $\tan \frac{x}{2}=-\frac{2}{3}$
• D. $\cos 2x=-\frac{7}{25}$

Answer 1

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

$\cos 2x=-\frac{7}{25}$

$\tan x=-\frac{4}{3}$
As $x$ lies in the second quadrant,
$\Rightarrow \cos x=-\frac{3}{5}$
Now,
$\frac{\pi }{2}<x<\pi \Rightarrow \frac{\pi }{4}<\frac{x}{2}<\frac{\pi }{2}$
So,
$\sin \frac{x}{2}=+\sqrt{\frac{1-\cos x}{2}}(\because \sin \frac{x}{2}>0)\Rightarrow \sin \frac{x}{2}=\sqrt{\frac{1-(-\frac{3}{5})}{2}}\Rightarrow \sin \frac{x}{2}=\frac{2}{\sqrt{5}}=\frac{2\sqrt{5}}{5}$
$\Rightarrow \cos \frac{x}{2}=\frac{\sqrt{25-20}}{5}=\frac{1}{\sqrt{5}}$
$\Rightarrow \tan \frac{x}{2}=2$

$\cos 2x=\frac{1-{\tan }^{2}x}{1+{\tan }^{2}x}\Rightarrow \cos 2x=\frac{1-\frac{16}{9}}{1+\frac{16}{9}}=-\frac{7}{25}$