MHT CET · Maths · Trigonometric Ratios & Identities
If \(\tan x=\frac{3}{4}\) and \(\pi \lt x \lt \frac{3 \pi}{2}\), then \(\cos \frac{x}{2}=\) \(\qquad\)
- A \(\frac{-2}{5}\)
- B \(\frac{2}{5}\)
- C \(\frac{1}{\sqrt{10}}\)
- D \(\frac{-1}{\sqrt{10}}\)
Answer & Solution
Correct Answer
(D) \(\frac{-1}{\sqrt{10}}\)
Step-by-step Solution
Detailed explanation
\(\tan x=\frac{3}{4}, \pi \lt x \lt \frac{3 \pi}{2} \)
\( \therefore 1+\tan ^2 x=\sec ^2 x \)
\( \therefore \sec ^2 x=\frac{25}{16} \)
\( \therefore \sec x=\frac{-5}{4} \ldots\left[\because \pi \lt x \lt \frac{3 \pi}{2}\right]\)
\(\therefore \cos x=\frac{-4}{5}\)
\(\therefore \cos \frac{x}{2}=-\sqrt{\frac{1+\cos x}{2}}\quad\)\(\ldots\left[\pi \lt x \lt \frac{3 \pi}{2} \Rightarrow \frac{\pi}{2} \lt \frac{x}{2} \lt \frac{3 \pi}{4}\right]\)
\(=-\sqrt{\frac{1}{10}}\)
\(=\frac{-1}{\sqrt{10}}\)
\( \therefore 1+\tan ^2 x=\sec ^2 x \)
\( \therefore \sec ^2 x=\frac{25}{16} \)
\( \therefore \sec x=\frac{-5}{4} \ldots\left[\because \pi \lt x \lt \frac{3 \pi}{2}\right]\)
\(\therefore \cos x=\frac{-4}{5}\)
\(\therefore \cos \frac{x}{2}=-\sqrt{\frac{1+\cos x}{2}}\quad\)\(\ldots\left[\pi \lt x \lt \frac{3 \pi}{2} \Rightarrow \frac{\pi}{2} \lt \frac{x}{2} \lt \frac{3 \pi}{4}\right]\)
\(=-\sqrt{\frac{1}{10}}\)
\(=\frac{-1}{\sqrt{10}}\)
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