JEE PYQ: Inverse Trigonometric Functions Question 15
Question 15 - 2019 (10 Apr Shift 2)
If $\cos^{-1} x - \cos^{-1}\frac{y}{2} = \alpha$, where $-1 \le x \le 1$, $-2 \le y \le 2$, $x \le \frac{y}{2}$, then for all $x, y$, $4x^2 - 4xy\cos\alpha + y^2$ is equal to:
(1) $4\sin^2\alpha$
(2) $2\sin^2\alpha$
(3) $4\sin^2\alpha - 2x^2y^2$
(4) $4\cos^2\alpha + 2x^2y^2$
Show Answer
Answer: (1)
Solution
$\cos^{-1}x - \cos^{-1}\frac{y}{2} = \alpha \Rightarrow \cos^{-1}\left(\frac{xy}{2} + \sqrt{1-x^2}\sqrt{1-\frac{y^2}{4}}\right) = \alpha$. So $\frac{xy}{2} + \sqrt{1-x^2}\sqrt{\frac{4-y^2}{4}} = \cos\alpha$. Then $(xy - 2\cos\alpha)^2 = (1-x^2)(4-y^2)$. Expanding: $x^2y^2 - 4xy\cos\alpha + 4\cos^2\alpha = 4 - y^2 - 4x^2 + x^2y^2$. So $4x^2 - 4xy\cos\alpha + y^2 = 4 - 4\cos^2\alpha = 4\sin^2\alpha$.
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