JEE PYQ: Inverse Trigonometric Functions Question 9
Question 9 - 2021 (26 Feb Shift 1)
If $\frac{\sin^{-1} x}{a} = \frac{\cos^{-1} x}{b} = \frac{\tan^{-1} y}{c}$, $0 < x < 1$, then the value of $\cos\left(\frac{\pi c}{a+b}\right)$ is:
(1) $\frac{1-y^2}{2y}$
(2) $\frac{1-y^2}{1+y^2}$
(3) $1 - y^2$
(4) $\frac{1-y^2}{y\sqrt{y}}$
Show Answer
Answer: (2)
Solution
$\frac{\sin^{-1}x}{a} = \frac{\cos^{-1}x}{b} = \frac{\sin^{-1}x + \cos^{-1}x}{a+b} = \frac{\pi}{2(a+b)}$. So $\frac{\tan^{-1}y}{c} = \frac{\pi}{2(a+b)}$, giving $2\tan^{-1}y = \frac{\pi c}{a+b}$. Therefore $\cos\left(\frac{\pi c}{a+b}\right) = \cos(2\tan^{-1}y) = \frac{1-y^2}{1+y^2}$.
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