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Inverse Trigonometric Functions

Section: Mathematics

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Question : 97 of 99

Marks: +1, -0

If

\cos^{-1} x - \cos^{-1} \;\frac{y}{2} = \alpha

4 x^2 - 4 x y \cos \alpha + y^2

$2 \sin 2 \alpha$
$4$
$4 \sin^2 \alpha$
$-4 \sin^2 \alpha$

Solution:

As we know,

\cos^{-1} A - \cos^{-1} B

= \cos^{-1} (AB + \sqrt{1-A^2} \cdot \sqrt{1-B^2})

Given,

\cos^{-1} x - \cos^{-1} \;\frac{y}{2} = \alpha

\Rightarrow \cos^{-1} \left( x \cdot \;\frac{y}{2} + \sqrt{1-x^2} \cdot \sqrt{1-\;\frac{y^2}{4}} \right) = \alpha

\Rightarrow \;\frac{xy}{2} + \sqrt{1-x^2} \sqrt{1-\;\frac{y^2}{4}} = \cos x

\Rightarrow \left(\cos x - \;\frac{xy}{2}\right)^2 = (1-x^2) \left(1-\;\frac{y^2}{4}\right)

\Rightarrow \cos^2 + \;\frac{x^2 y^2}{4} - 2 \cdot \cos x \cdot \;\frac{xy}{2}

= 1 - x^2 - \;\frac{y^2}{4} + \;\frac{x^2 y^2}{4}

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

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

\Rightarrow 4 x^2 + y^2 - 4 x y \cos x = 4 \sin^2 x

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