JEE Main 2022 (Online) 30th June Morning ShiftQuestion 1
Let $\alpha=\tan\left(\frac{5\pi}{16}\sin\left(2\cos^{-1}\left(\frac{1}{\sqrt{5}}\right)\right)\right)$ and $\beta=\cos\left(\sin^{-1}\left(\frac{4}{5}\right)+\sec^{-1}\left(\frac{5}{3}\right)\right)$ where the inverse trigonometric functions take principal values. Then, the equation whose roots are $\alpha$ and $\beta$ is :
Inverse Trigonometric Functions
Solution
$\cos^{-1}\frac1{\sqrt5}=\theta$: $\sin2\theta=2\cdot\frac2{\sqrt5}\cdot\frac1{\sqrt5}=\frac45$, so $\alpha=\tan(\frac{5\pi}{16}\cdot\frac45)=\tan\frac\pi4=1$. $\sec^{-1}\frac53=\cos^{-1}\frac35=\phi$, and $\sin^{-1}\frac45=\phi$ too (since $\sin\phi=\frac45$). $\beta=\cos2\phi=\frac9{25}-\frac{16}{25}=-\frac7{25}$. $\alpha+\beta=\frac{18}{25}$, $\alpha\beta=-\frac7{25}$, so $25x^2-18x-7=0$. Option (C).