JEE Mains · Maths · STD 12 - 2. inverse trigonometric function
If \(y=\cos \left(\frac{\pi}{3}+\cos ^{-1} \frac{x}{2}\right)\), then \((x-y)^2+3 y^2\) is equal to _____.
- A 1
- B 2
- C 3
- D 4
Answer & Solution
Correct Answer
(C) 3
Step-by-step Solution
Detailed explanation
\begin{aligned} & y=\cos \left(\cos ^{-1} \frac{1}{2}+\cos ^{-1} \frac{x}{2}\right) \\ & y=\frac{1}{2} \times \frac{x}{2}-\sqrt{1-\frac{1}{4}} \sqrt{1-\frac{x^2}{4}} \\ & 4 y=x-\sqrt{3} \sqrt{4-x^2} \\ & 3\left(4-x^2\right)=x^2+16 y^2-8 x y \\ & 12-3 x^2=x^2+16 y^2-8 x y \\ & 4…
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