Question

If $\tan x=\frac{3}{4},\pi <x<\frac{3\pi }{2}$ find value of $\sin \frac{x}{2},\cos \frac{x}{2},\tan \frac{x}{2}$

• A. cos x/2= 3/2, sin x/2 =-3/2, tan x/2= -1/√10
• B. cos x/2= 3/√10, sin x/2 =-3/2, tan x/2= -1/√10
• C. cos x/2= -1/√10, sin x/2 = 3/√10, tan x/2= -3
• D. cos x/2= 3/√10, sin x/2 =-2/3, tan x/2= -1/√10

Answer 1

The correct option is C cos x/2= -1/√10, sin x/2 = 3/√10, tan x/2= -3

Given

$\pi <x<\frac{3\pi }{2}$

$⟹\frac{\pi }{2}<\frac{x}{2}<\frac{3\pi }{4}$

$⟹\cos \frac{x}{2}<0,\sin \frac{x}{2}>0$

$\tan x=\frac{3}{4}$

$\frac{2\tan \frac{x}{2}}{1-{\tan }^2\frac{x}{2}}=\frac{3}{4}$

$8\tan \frac{x}{2}=3-3{\tan }^2\frac{x}{2}$

$3{\tan }^2\frac{x}{2}+8\tan \frac{x}{2}-3=0$

$3{\tan }^2\frac{x}{2}+9\tan \frac{x}{2}-(\tan \frac{x}{2}+3)=0$

$\tan \frac{x}{2}=-3,\frac{1}{3}$

As $tan\frac{x}{2}<0⟹\tan \frac{x}{2}=-3$

$\cos \frac{x}{2}=\frac{-1}{\sqrt{(-1{)}^2+3^2}}=\frac{-1}{\sqrt{10}}$

$\sin \frac{x}{2}=\frac{3}{\sqrt{(-1{)}^2+3^2}}=\frac{3}{\sqrt{10}}$