JEE Advanced 2022 Paper 1 · Q01 · Inverse Trigonometric Functions

Question

Considering only the principal values of the inverse trigonometric functions, the value of $\dfrac{3}{2}\cos^{-1}\sqrt{\dfrac{2}{2+\pi^{2}}}+\dfrac{1}{4}\sin^{-1}\dfrac{2\sqrt{2}\,\pi}{2+\pi^{2}}+	an^{-1}\dfrac{\sqrt{2}}{\pi}$ is __________ .

SolveKar AI · Accepted Answer

Let $	heta=	an^{-1}\dfrac{\sqrt{2}}{\pi}\in(0,\pi/2)$. Then $	an	heta=\sqrt{2}/\pi$, so $\sin	heta=\sqrt{2}/\sqrt{2+\pi^{2}}$, $\cos	heta=\pi/\sqrt{2+\pi^{2}}$. Hence $\cos 2	heta=\dfrac{\pi^{2}-2}{\pi^{2}+2}$ and $\sin 2	heta=\dfrac{2\sqrt{2}\pi}{\pi^{2}+2}$. Also $\cos^{-1}\sqrt{\dfrac{2}{2+\pi^{2}}}=\cos^{-1}(\sin	heta)=\dfrac{\pi}{2}-	heta$. So the expression $=\dfrac{3}{2}\!\left(\dfrac{\pi}{2}-	hetaight)+\dfrac{1}{4}(2	heta)+	heta=\dfrac{3\pi}{4}-\dfrac{3	heta}{2}+\dfrac{	heta}{2}+	heta=\dfrac{3\pi}{4}$. Numerically $3\pi/4\approx 2.356\Rightarrow 2.36$ (the key also accepts 2.35).

JEE Advanced 2022 Paper 1 · Q01 · Inverse Trigonometric Functions

Solution