Question

If $\tan \left(x\right)=\frac{3}{4},\pi <x<\frac{3\pi }{2}$, then the value of $\cos \left(\frac{x}{2}\right)$ is

• A. $-\frac{1}{\sqrt{10}}$
• B. $\frac{3}{\sqrt{10}}$
• C. $\frac{1}{\sqrt{10}}$
• D. $-\frac{3}{\sqrt{10}}$

Answer 1

The correct option is A

$-\frac{1}{\sqrt{10}}$

Explanation for correct option

Step 1. Find the value of $\cos \left(x\right)$

Given that $\tan \left(x\right)=\frac{3}{4},\pi <x<\frac{3\pi }{2}$

We know that ${\sec }^2\left(x\right)-{\tan }^2\left(x\right)=1$

We know that $\mathrm{cosine}$and $\mathrm{secant}$ is negative in third quadrant

$$\begin{gathered} \therefore \sec \left(x\right)=-\sqrt{1+{\tan }^2\left(x\right)}\left[\because \pi <x<\frac{3\pi }{2}\right] \\ \Rightarrow \sec \left(x\right)=-\sqrt{1+{\left(\frac{3}{4}\right)}^2} \\ \Rightarrow \sec \left(x\right)=-\frac{5}{4} \\ \Rightarrow \sec \left(x\right)=-\frac{5}{4} \\ \iff \cos \left(x\right)=-\frac{4}{5}\left[\because \cos \left(x\right)=\frac{1}{\sec \left(x\right)}\right] \end{gathered}$$

Step 2. Find the value of $\cos \left(\frac{x}{2}\right)$

$\cos \left(x\right)=-\frac{4}{5}$

We know that

$$\begin{gathered} \cos \left(x\right)={\cos }^2\left(\frac{x}{2}\right)-{\sin }^2\left(\frac{x}{2}\right) \\ \Rightarrow \cos \left(x\right)=2{\cos }^2\left(\frac{x}{2}\right)-1\left[\because {\cos }^2\left(x\right)+{\sin }^2\left(x\right)=1\right] \\ \Rightarrow {\cos }^2\left(\frac{x}{2}\right)=\frac{\cos \left(x\right)+1}{2} \\ \Rightarrow \cos \left(\frac{x}{2}\right)=\pm \sqrt{\frac{\cos \left(x\right)+1}{2}} \\ \Rightarrow \cos \left(\frac{x}{2}\right)=\pm \sqrt{\frac{-{\displaystyle \frac{4}{5}}+1}{2}} \\ \Rightarrow \cos \left(\frac{x}{2}\right)=\pm \frac{1}{\sqrt{10}} \\ \because \pi <x<\frac{3\pi }{2} \\ \therefore \frac{\pi }{2}<x<\frac{3\pi }{4} \end{gathered}$$

That is the second quadrant and in second quadrant $\mathrm{cosine}$ is negative

$\therefore \cos \left(\frac{x}{2}\right)=-\frac{1}{\sqrt{10}}$

Hence, option A is correct