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(A) Let $x = \frac{\pi}{\sqrt{2}}$. Then $	an ^{-1} x = 	an ^{-1} \frac{\pi}{\sqrt{2}}$.From the given triangle,$\cos 	heta = \frac{\sqrt{2}}{\sqrt

Vedclass · Accepted Answer

$2.35$

Vedclass · Answer

(A) Let $x = \frac{\pi}{\sqrt{2}}$. Then $	an ^{-1} x = 	an ^{-1} \frac{\pi}{\sqrt{2}}$.From the given triangle,$\cos 	heta = \frac{\sqrt{2}}{\sqrt{2+\pi^2}}$,so $\cos ^{-1} \sqrt{\frac{2}{2+\pi^2}} = 	an ^{-1} \frac{\pi}{\sqrt{2}}$.Next,consider $\sin ^{-1} \left( \frac{2 \sqrt{2} \pi}{2+\pi^2} ight) = \sin ^{-1} \left( \frac{2(\pi/\sqrt{2})}{1+(\pi/\sqrt{2})^2} ight)$.Since $x = \frac{\pi}{\sqrt{2}} \approx \frac{3.14}{1.414} > 1$,we use the formula $\sin ^{-1} \left( \frac{2x}{1+x^2} ight) = \pi - 2 	an ^{-1} x$.Thus,$\sin ^{-1} \left( \frac{2 \sqrt{2} \pi}{2+\pi^2} ight) = \pi - 2 	an ^{-1} \frac{\pi}{\sqrt{2}}$.Also,$	an ^{-1} \frac{\sqrt{2}}{\pi} = \cot ^{-1} \frac{\pi}{\sqrt{2}}$.Substituting these into the expression:Expression $= \frac{3}{2} 	an ^{-1} \frac{\pi}{\sqrt{2}} + \frac{1}{4} \left( \pi - 2 	an ^{-1} \frac{\pi}{\sqrt{2}} ight) + \cot ^{-1} \frac{\pi}{\sqrt{2}}$.$= \left( \frac{3}{2} - \frac{1}{2} ight) 	an ^{-1} \frac{\pi}{\sqrt{2}} + \frac{\pi}{4} + \cot ^{-1} \frac{\pi}{\sqrt{2}}$.$= 	an ^{-1} \frac{\pi}{\sqrt{2}} + \cot ^{-1} \frac{\pi}{\sqrt{2}} + \frac{\pi}{4}$.$= \frac{\pi}{2} + \frac{\pi}{4} = \frac{3\pi}{4}$.Using $\pi \approx 3.14159$,$\frac{3 	imes 3.14159}{4} \approx 2.356$. Thus,the value is approximately $2.35$.

Considering only the principal values of the inverse trigonometric functions,the value of $\frac{3}{2} \cos ^{-1} \sqrt{\frac{2}{2+\pi^2}}+\frac{1}{4} \sin ^{-1} \frac{2 \sqrt{2} \pi}{2+\pi^2}+\tan ^{-1} \frac{\sqrt{2}}{\pi}$ is:

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