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

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(D) આપણે $|x| < 1$ માટે $2 	an ^{-1}(x) = 	an ^{-1}\left(\frac{2x}{1-x^2}ight)$ સૂત્રનો ઉપયોગ કરીએ છીએ.પ્રથમ,$2 	an ^{-1}\left(\frac{1}{3}ight)

Vedclass · Accepted Answer

$\frac{\pi}{2}$

Vedclass · Answer

(D) આપણે $|x| પ્રથમ,$2 	an ^{-1}\left(\frac{1}{3}ight) = 	an ^{-1}\left(\frac{2 	imes \frac{1}{3}}{1-(\frac{1}{3})^2}ight) = 	an ^{-1}\left(\frac{2/3}{8/9}ight) = 	an ^{-1}\left(\frac{3}{4}ight)$ ગણો.હવે,$\cos ^{-1}\left(\frac{3}{5}ight)$ ને $	an ^{-1}$ માં ફેરવો. ધારો કે $	heta = \cos ^{-1}\left(\frac{3}{5}ight)$,તો $\cos 	heta = \frac{3}{5}$. $\cos^2 	heta + \sin^2 	heta = 1$ હોવાથી,$\sin 	heta = \sqrt{1 - (3/5)^2} = 4/5$. તેથી,$	an 	heta = \frac{4/5}{3/5} = \frac{4}{3}$,એટલે કે $\cos ^{-1}\left(\frac{3}{5}ight) = 	an ^{-1}\left(\frac{4}{3}ight)$.પદાવલિ $	an ^{-1}\left(\frac{3}{4}ight) + 	an ^{-1}\left(\frac{4}{3}ight)$ બને છે.$x > 0$ માટે $	an ^{-1}(x) + 	an ^{-1}(1/x) = \frac{\pi}{2}$ નો ઉપયોગ કરતા,આપણને $	an ^{-1}\left(\frac{3}{4}ight) + 	an ^{-1}\left(\frac{1}{3/4}ight) = \frac{\pi}{2}$ મળે છે.

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

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