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
- A \(\frac{\pi}{6}\)
- B \(\frac{\pi}{3}\)
- C \(\pi\)
- D \(0\)
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…
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