If {^{-1}(x)}{a}={^{-1} x}{b}={^{-1} y}{c}, 0 [26 Feb 2021 Shift 1]

Correct Answer is : 1−y21+y2{1-y²}{1+y²}1+y21−y2

Correct Answer is : 1−y21+y2{1-y²}{1+y²}1+y21−y2

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JEE Main 26 Feb 2021 Shift 1 Solved Paper

Section: Mathematics

Question : 72 of 90

Marks: +1, -0

\frac{\sin^{-1}(x)}{a}=\frac{\cos^{-1} x}{b}=\frac{\tan^{-1} y}{c}, 0 < x < 1

\cos \left(\frac{\pi c}{a+b}\right)

Solution:

\frac{\sin^{-1} x}{a}=\frac{\cos^{-1} x}{b}=\frac{\tan^{-1} y}{c}

Take first two terms of Eq. (i)

\frac{\sin^{-1} x}{a}=\frac{\cos^{-1} x}{b}

\Rightarrow \frac{\sin^{-1} x}{a}=\frac{\cos^{-1} x}{b}=\frac{\sin^{-1} x+\cos^{-1} x}{a+b}

\Rightarrow \frac{\sin^{-1} x}{a}=\frac{\cos^{-1} x}{b}=\frac{\frac{\pi}{2}}{a+b} \left[ \because \sin^{-1} x+\cos^{-1} x=\frac{\pi}{2} \right]

\Rightarrow \frac{\sin^{-} x}{a}=\frac{\cos^{-1} x}{b}=\frac{\frac{\pi}{2}}{a+b}=\frac{\tan^{-1} y}{c}

Using last two terms,

\frac{\tan^{-1} y}{c} = \frac{\frac{\pi}{2}}{a+b}

\Rightarrow \tan^{-1} y = \frac{\pi c}{2(a+b)}

\Rightarrow 2 \tan^{-1} y = \frac{\pi c}{(a+b)}

\Rightarrow \cos^{-1} \left( \frac{1-y^{2}}{1+y^{2}} \right) = \frac{\pi c}{a+b}

\left[ \because 2\tan^{-1} y = \cos^{-1} \left( \frac{1-y^{2}}{1+y^{2}} \right) \right]

\Rightarrow \frac{1-y^{2}}{1+y^{2}} = \cos \left( \frac{\pi c}{a+b} \right)

\therefore \cos \left( \frac{\pi c}{a+b} \right) = \frac{1-y^{2}}{1+y^{2}}

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