operatorname{Tan}^{-1} frac{3}{5} + operatorn | vedclass

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Question

(A) આપણે સૂત્ર $\operatorname{Tan}^{-1} x + \operatorname{Tan}^{-1} y = \operatorname{Tan}^{-1} \left( \frac{x+y}{1-xy} \right)$ નો ઉપયોગ કરીએ છીએ.પ્ર

Vedclass · Accepted Answer

$\operatorname{Tan}^{-1} \frac{9}{10}$

Vedclass · Answer

(A) આપણે સૂત્ર $\operatorname{Tan}^{-1} x + \operatorname{Tan}^{-1} y = \operatorname{Tan}^{-1} \left( \frac{x+y}{1-xy} ight)$ નો ઉપયોગ કરીએ છીએ.પ્રથમ,$\operatorname{Tan}^{-1} \frac{3}{5} + \operatorname{Tan}^{-1} \frac{6}{41}$ ની ગણતરી કરો:$= \operatorname{Tan}^{-1} \left( \frac{\frac{3}{5} + \frac{6}{41}}{1 - \frac{3}{5} 	imes \frac{6}{41}} ight) = \operatorname{Tan}^{-1} \left( \frac{\frac{123+30}{205}}{\frac{205-18}{205}} ight) = \operatorname{Tan}^{-1} \left( \frac{153}{187} ight) = \operatorname{Tan}^{-1} \left( \frac{9}{11} ight)$.હવે,ત્રીજું પદ ઉમેરો: $\operatorname{Tan}^{-1} \frac{9}{11} + \operatorname{Tan}^{-1} \frac{9}{191}$:$= \operatorname{Tan}^{-1} \left( \frac{\frac{9}{11} + \frac{9}{191}}{1 - \frac{9}{11} 	imes \frac{9}{191}} ight) = \operatorname{Tan}^{-1} \left( \frac{\frac{1719+99}{2101}}{\frac{2101-81}{2101}} ight) = \operatorname{Tan}^{-1} \left( \frac{1818}{2020} ight) = \operatorname{Tan}^{-1} \left( \frac{9}{10} ight)$.

$\operatorname{Tan}^{-1} \frac{3}{5} + \operatorname{Tan}^{-1} \frac{6}{41} + \operatorname{Tan}^{-1} \frac{9}{191} = $

Similar Questions

સાબિત કરો કે $\tan ^{-1} \sqrt{x} = \frac{1}{2} \cos ^{-1} \left( \frac{1-x}{1+x} \right)$,જ્યાં $x \in [0, 1]$.

$x$ ની સાપેક્ષમાં $\cos^{-1} \left( \sqrt{\frac{1+x}{2}} \right)$ નું વિકલન સહગુણક શું છે?

$ \cos \left[2 \sin ^{-1} \frac{3}{4} + \cos ^{-1} \frac{3}{4}\right] $

$\tan ^{-1}\left(\frac{1}{3}\right)+\tan ^{-1}\left(\frac{1}{5}\right)+\tan ^{-1}\left(\frac{1}{7}\right)+\tan ^{-1}\left(\frac{1}{8}\right)$ ની કિંમત શોધો.

જો $x=\operatorname{cosec}(\tan ^{-1}(\cos (\cot ^{-1}(\sec (\sin ^{-1} a)))))$,જ્યાં $a \in [0, 1]$,તો નીચેનામાંથી કયું સાચું છે?

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