Q96 / 180 · KCET 2014 · Mathematics
If \( \tan x=\frac{3}{4}, \Pi < x < \frac{3 \Pi}{2} \), then the value of \( \cos \frac{x}{2} \) is
- A\( \frac{3}{\sqrt{10}} \)
- B\( -\frac{3}{\sqrt{10}} \)
- Correct answerC\( -\frac{1}{\sqrt{10}} \)
- D\( \frac{1}{\sqrt{10}} \)
Solution
Given,
\( \tan x=\frac{3}{4}, \Pi < x < 3 \frac{\Pi}{2} \) \( \Rightarrow \frac{\Pi}{2} < \frac{x}{2} < 3 \frac{\Pi}{4} \) (belongs to quadrant II)
In the second quadrant, \( \cos \left(\frac{x}{2}\right) \) is negative.
\[ \begin{array}{l} \sec 2 x=1+\tan 2 x=1+\left(\frac{3}{4}\right) 2=\frac{(16+9)}{16}=\frac{25}{16} \\ \sec x=\pm \frac{5}{4} \end{array} \]
Thus, \( \cos x=-\frac{4}{5}\left\{\right. \) since \( \left.\Pi < x < 3 \frac{\Pi}{2}\right\} \)
Using the formula, \( \cos 2 A=2 \cos 2 A-1 \),
\( 2 \cos 2\left(\frac{x}{2}\right)=1+\cos x=1+\left(-\frac{4}{5}\right) \)
\( 2 \cos 2\left(\frac{x}{2}\right)=\frac{1}{5} \)
\( \cos 2\left(\frac{x}{2}\right)=\frac{1}{10} \)
\( \cos \left(\frac{x}{2}\right)=\pm \frac{\sqrt{1}}{10}=-\frac{1}{\sqrt{10}}\left\{\right. \) since \( \frac{x}{2} \) lies in the quadrant II\}
\( \tan x=\frac{3}{4}, \Pi < x < 3 \frac{\Pi}{2} \) \( \Rightarrow \frac{\Pi}{2} < \frac{x}{2} < 3 \frac{\Pi}{4} \) (belongs to quadrant II)
In the second quadrant, \( \cos \left(\frac{x}{2}\right) \) is negative.
\[ \begin{array}{l} \sec 2 x=1+\tan 2 x=1+\left(\frac{3}{4}\right) 2=\frac{(16+9)}{16}=\frac{25}{16} \\ \sec x=\pm \frac{5}{4} \end{array} \]
Thus, \( \cos x=-\frac{4}{5}\left\{\right. \) since \( \left.\Pi < x < 3 \frac{\Pi}{2}\right\} \)
Using the formula, \( \cos 2 A=2 \cos 2 A-1 \),
\( 2 \cos 2\left(\frac{x}{2}\right)=1+\cos x=1+\left(-\frac{4}{5}\right) \)
\( 2 \cos 2\left(\frac{x}{2}\right)=\frac{1}{5} \)
\( \cos 2\left(\frac{x}{2}\right)=\frac{1}{10} \)
\( \cos \left(\frac{x}{2}\right)=\pm \frac{\sqrt{1}}{10}=-\frac{1}{\sqrt{10}}\left\{\right. \) since \( \frac{x}{2} \) lies in the quadrant II\}
Hence, the correct answer is option: 'C'.