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

જો $x \in \left(0, \frac{1}{4}\right)$ માટે,$\tan^{-1}\left(\frac{6x\sqrt{x}}{1-9x^3}\right)$ નું વિકલન $\sqrt{x} \cdot g(x)$ હોય,તો $g(x)$ ની કિંમત શોધો.

$2{\tan ^{ - 1}}\left[ {\sqrt {\frac{{a - b}}{{a + b}}} \tan \frac{\theta }{2}} \right] = $

જો $\sin \left(\cot ^{-1}(x+1)\right)=\cos \left(\tan ^{-1} x\right)$ હોય,તો $x$ ની કિંમત કેટલી થાય?

$\sin(3 \sin^{-1}(1/5))$ ની કિંમત શોધો.

જો $0 \leqslant \cos ^{-1} x \leqslant \pi$ અને $\frac{-\pi}{2} \leqslant \sin ^{-1} x \leqslant \frac{\pi}{2}$ હોય,તો $x=\frac{1}{5}$ માટે $\cos \left(2 \cos ^{-1} x+\sin ^{-1} x\right)$ ની કિંમત શોધો.

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