SATHEE: Inverse Circular Functions 3 Question 2

Inverse Circular Functions 3 Question 2

2. If $\cos^{- 1} x - \cos^{- 1} \frac{y}{2} = α$ , where $- 1 \leq x \leq 1, - 2 \leq y \leq 2$ , $x \leq \frac{y}{2}$ , then for all $x, y, 4 x^{2} - 4 x y \cos α + y^{2}$ is equal to

(a) $2 \sin^{2} α$

(b) $4 \cos^{2} α + 2 x^{2} y^{2}$

(d) $4 \sin^{2} α - 2 x^{2} y^{2}$

(2019 Main, 10 April II)

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Answer:

Correct Answer: 2. (c)

Solution:

Given equation is

$\begin{aligned} \cos^{- 1} x - \cos^{- 1} \frac{y}{2} = α, where - 1 \leq x \leq 1, \\ - 2 \leq y \leq 2 and x \leq \frac{y}{2} \\ \cos^{- 1} x \frac{y}{2} + \sqrt{1 - x^{2}} \sqrt{1 - (y / 2)^{2}} = α \\ [∵ \cos^{- 1} x - \cos^{- 1} y = \cos^{- 1} (x y + \sqrt{1 - x^{2}} \sqrt{1 - y^{2})}, \\ \Rightarrow \frac{x y}{2} + \sqrt{1 - x^{2}} \sqrt{1 - (y / 2)^{2}} = \cos α \\ \Rightarrow \sqrt{1 - x^{2}} \sqrt{1 - (y / 2)^{2}} = \cos α - \frac{x y}{2} \end{aligned}$

On squaring both sides, we get

$(1 - x^{2}) 1 - \frac{y^{2}}{4} = \cos^{2} α + \frac{x^{2} y^{2}}{4} - 2 \frac{x y}{2} \cos α$

$\Rightarrow 1 - x^{2} - \frac{y^{2}}{4} + \frac{x^{2} y^{2}}{4} = \cos^{2} α + \frac{x^{2} y^{2}}{4} - x y \cos α$

$\Rightarrow x^{2} + \frac{y^{2}}{4} - x y \cos α = 1 - \cos^{2} α$

$\Rightarrow 4 x^{2} - 4 x y \cos α + y^{2} = 4 \sin^{2} α$

Inverse Circular Functions 3 Question 2

2. If $\cos^{- 1} x - \cos^{- 1} \frac{y}{2} = α$ , where $- 1 \leq x \leq 1, - 2 \leq y \leq 2$ , $x \leq \frac{y}{2}$ , then for all $x, y, 4 x^{2} - 4 x y \cos α + y^{2}$ is equal to

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Inverse Circular Functions 3 Question 2

2. If cos−1⁡x−cos−1⁡y2=α, where −1≤x≤1,−2≤y≤2, x≤y2, then for all x,y,4x2−4xycos⁡α+y2 is equal to

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2. If $\cos^{- 1} x - \cos^{- 1} \frac{y}{2} = α$ , where $- 1 \leq x \leq 1, - 2 \leq y \leq 2$ , $x \leq \frac{y}{2}$ , then for all $x, y, 4 x^{2} - 4 x y \cos α + y^{2}$ is equal to