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MHT CET · Maths · Inverse Trigonometric Functions

The value of \(\cos ^{-1}\left\{\frac{1}{\sqrt{2}}\left(\cos \frac{9 \pi}{10}-\sin \frac{9 \pi}{10}\right)\right\}\) is

  1. A \(\frac{7 \pi}{20}\)
  2. B \(\frac{13 \pi}{20}\)
  3. C \(\frac{17 \pi}{20}\)
  4. D \(\frac{21 \pi}{20}\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(\frac{17 \pi}{20}\)

Step-by-step Solution

Detailed explanation

\(\begin{aligned} & \cos ^{-1}\left\{\frac{1}{\sqrt{2}}\left[\cos \left(\frac{9 \pi}{10}\right)-\sin \left(\frac{9 \pi}{10}\right)\right]\right\} \\ & =\cos ^{-1}\left\{\frac{1}{\sqrt{2}}\left(\cos \left(\pi-\frac{\pi}{10}\right)-\sin \left(\pi-\frac{\pi}{10}\right)\right)\right\} \\ & =\cos ^{-1}\left\{\frac{1}{\sqrt{2}}\left[-\cos \left(\frac{\pi}{10}\right)-\sin \left(\frac{\pi}{10}\right)\right]\right\}\end{aligned}\)
\(\begin{aligned} & =\cos ^{-1}\left\{(-1)\left[\frac{1}{\sqrt{2}} \cos \left(\frac{\pi}{10}\right)+\frac{1}{\sqrt{2}} \sin \left(\frac{\pi}{10}\right)\right]\right\} \\ & =\cos ^{-1}\left\{(-1)\left[\cos \frac{\pi}{4} \cos \frac{\pi}{10}+\sin \frac{\pi}{4} \sin \frac{\pi}{10}\right]\right\}\end{aligned}\)
\(\begin{aligned} & =\cos ^{-1}\left\{(-1)\left[\cos \left(\frac{\pi}{4}-\frac{\pi}{10}\right)\right]\right\} \\ & =\cos ^{-1}\left\{(-1)\left[\cos \left(\frac{3 \pi}{20}\right)\right]\right\} \\ & =\pi-\cos ^{-1}\left(\cos \frac{3 \pi}{20}\right) \\ & =\pi-\frac{3 \pi}{20} \\ & =\frac{17 \pi}{20}\end{aligned}\)