કિંમત શોધો: tan^{-1}left(frac{1}{4}right) + t | vedclass

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Question

(A) સૂત્ર $	an^{-1}x + 	an^{-1}y = 	an^{-1}\left(\frac{x+y}{1-xy}ight)$ નો ઉપયોગ કરતા:$	an^{-1}\left(\frac{1}{4}ight) + 	an^{-1}\left(\frac{2

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) સૂત્ર $	an^{-1}x + 	an^{-1}y = 	an^{-1}\left(\frac{x+y}{1-xy}ight)$ નો ઉપયોગ કરતા:$	an^{-1}\left(\frac{1}{4}ight) + 	an^{-1}\left(\frac{2}{9}ight) = 	an^{-1}\left(\frac{\frac{1}{4} + \frac{2}{9}}{1 - \frac{1}{4} 	imes \frac{2}{9}}ight)$$= 	an^{-1}\left(\frac{\frac{9+8}{36}}{1 - \frac{2}{36}}ight) = 	an^{-1}\left(\frac{17/36}{34/36}ight) = 	an^{-1}\left(\frac{17}{34}ight) = 	an^{-1}\left(\frac{1}{2}ight)$હવે,આપણે સૂત્ર $2	an^{-1}x = \cos^{-1}\left(\frac{1-x^2}{1+x^2}ight)$ નો ઉપયોગ કરીએ:$	an^{-1}\left(\frac{1}{2}ight) = \frac{1}{2} \left( 2	an^{-1}\left(\frac{1}{2}ight) ight) = \frac{1}{2} \cos^{-1}\left(\frac{1-(1/2)^2}{1+(1/2)^2}ight)$$= \frac{1}{2} \cos^{-1}\left(\frac{1-1/4}{1+1/4}ight) = \frac{1}{2} \cos^{-1}\left(\frac{3/4}{5/4}ight) = \frac{1}{2} \cos^{-1}\left(\frac{3}{5}ight)$આમ,સાચો વિકલ્પ $A$ છે.

કિંમત શોધો: $\tan^{-1}\left(\frac{1}{4}\right) + \tan^{-1}\left(\frac{2}{9}\right) = $

Similar Questions

$\tan \left[ \sin^{-1} \left( \frac{3}{5} \right) + \cos^{-1} \left( \frac{3}{\sqrt{13}} \right) \right]$ ની કિંમત શોધો.

જો $\tan ^{-1}x + \tan ^{-1}y + \tan ^{-1}z = \frac{\pi }{2}$ હોય,તો

જો $\frac{a}{b} \tan x > -1$ હોય,તો $\tan ^{-1}\left[\frac{a \cos x-b \sin x}{b \cos x+a \sin x}\right]$ નું સાદું રૂપ આપો.

$\cos \left(2 \cos ^{-1} \frac{1}{5}+\sin ^{-1} \frac{1}{5}\right)$ ની કિંમત શોધો.

જો $(\tan ^{-1} x)^2+(\cot ^{-1} x)^2=\frac{5 \pi^2}{8}$ હોય,તો $x^2+1=$

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