2 tan ^{-1}left(frac{1}{3}right)+cos ^{-1}lef | vedclass

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(D) We use the formula $2 	an ^{-1}(x) = 	an ^{-1}\left(\frac{2x}{1-x^2}ight)$ for $|x| < 1$.First,calculate $2 	an ^{-1}\left(\frac{1}{3}ight)

Vedclass · Accepted Answer

$\frac{\pi}{2}$

Vedclass · Answer

(D) We use the formula $2 	an ^{-1}(x) = 	an ^{-1}\left(\frac{2x}{1-x^2}ight)$ for $|x| First,calculate $2 	an ^{-1}\left(\frac{1}{3}ight) = 	an ^{-1}\left(\frac{2 	imes \frac{1}{3}}{1-(\frac{1}{3})^2}ight) = 	an ^{-1}\left(\frac{2/3}{8/9}ight) = 	an ^{-1}\left(\frac{3}{4}ight)$.Now,convert $\cos ^{-1}\left(\frac{3}{5}ight)$ to $	an ^{-1}$. Let $	heta = \cos ^{-1}\left(\frac{3}{5}ight)$,then $\cos 	heta = \frac{3}{5}$. Since $\cos^2 	heta + \sin^2 	heta = 1$,$\sin 	heta = \sqrt{1 - (3/5)^2} = 4/5$. Thus,$	an 	heta = \frac{4/5}{3/5} = \frac{4}{3}$,so $\cos ^{-1}\left(\frac{3}{5}ight) = 	an ^{-1}\left(\frac{4}{3}ight)$.The expression becomes $	an ^{-1}\left(\frac{3}{4}ight) + 	an ^{-1}\left(\frac{4}{3}ight)$.Using $	an ^{-1}(x) + 	an ^{-1}(1/x) = \frac{\pi}{2}$ for $x > 0$,we get $	an ^{-1}\left(\frac{3}{4}ight) + 	an ^{-1}\left(\frac{1}{3/4}ight) = \frac{\pi}{2}$.

$2 \tan ^{-1}\left(\frac{1}{3}\right)+\cos ^{-1}\left(\frac{3}{5}\right)=$

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