Question

Find the value of $\cos \left(\frac{x}{2}\right)$, if $\tan \left(x\right)=\frac{5}{12}$ and $x$ lies in quadrant $3.$

• A. $\frac{5}{\sqrt{13}}$
• B. $\frac{5}{\sqrt{26}}$
• C. $\frac{5}{2\sqrt{13}}$
• D. $\frac{-1}{\sqrt{26}}$

Answer 1

The correct option is D

$\frac{-1}{\sqrt{26}}$

Step:1 Calculate the value of $\sec \left(x\right)$ using Trigonometric identities

$$\begin{gathered} \tan \left(x\right)=\frac{5}{12} \\ (180°<x<270°) \\ \Rightarrow 90°<\frac{x}{2}<135° \\ {\sec }^2\left(x\right)=1+{\tan }^2\left(x\right) \\ =1+{\left(\frac{5}{12}\right)}^2 \\ =1+\frac{25}{144} \\ =\frac{169}{144} \\ \therefore \sec \left(x\right)=\pm \sqrt{\frac{169}{144}} \\ =-\frac{13}{12} \end{gathered}$$

In third quadrant, $\sec \left(x\right),\cos \left(x\right)$ are negative.

Step:2 Calculate the values of $\cos \left(x\right),\cos \left(\frac{x}{2}\right)$ using Trigonometric identities

$\therefore \cos \left(x\right)=-\frac{12}{13}$

We know, $$\begin{gathered} 2{\cos }^2\left(\frac{x}{2}\right)=1+\cos \left(x\right) \\ \Rightarrow 2{\cos }^2\left(\frac{x}{2}\right)=1-\frac{12}{13} \\ \Rightarrow 2{\cos }^2\left(\frac{x}{2}\right)=\frac{1}{13} \\ \Rightarrow {\cos }^2\left(\frac{x}{2}\right)=\frac{1}{26} \\ \Rightarrow \cos \left(\frac{x}{2}\right)=-\frac{1}{\sqrt{26}}[90°<\cos \left(\frac{x}{2}\right)<135°,\cos \left(\frac{x}{2}\right)=-ve] \end{gathered}$$

Therefore, the correct option is (D) i.e. $-\frac{1}{\sqrt{26}}$.