ExamBro
ExamBro
TS EAMCET · Maths · Inverse Trigonometric Functions

If \(\frac{1}{2} \leq x \leq 1\), then \(\cos ^{-1} x+\cos ^{-1}\left(\frac{x}{2}+\frac{1}{2} \sqrt{3-3 x^2}\right)\) is equal to

  1. A \(\frac{\pi}{6}\)
  2. B \(\frac{\pi}{3}\)
  3. C \(\pi\)
  4. D \(0\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(\frac{\pi}{3}\)

Step-by-step Solution

Detailed explanation

\begin{aligned} & \therefore \cos ^{-1} x+\cos ^{-1}\left(\frac{x}{2}+\frac{\sqrt{3}}{2} \sqrt{1-x^2}\right) \\ & =\cos ^{-1} x+\cos ^{-1}\left(x \times \frac{1}{2}+\frac{\sqrt{3}}{2} \times \sqrt{1-x^2}\right) \\ & =\cos ^{-1} x+\cos ^{-1}\left(\frac{1}{2}\right)-\cos ^{-1} x…