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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) આપણે સૂત્ર $2 	an ^{-1} x = 	an ^{-1} \left( \frac{2x}{1-x^2} ight)$ નો ઉપયોગ કરીએ છીએ.$x = \frac{1}{3}$ માટે,આપણને મળે છે:$2 	an ^{-1} \left

Vedclass · Accepted Answer

$0$

Vedclass · Answer

(A) આપણે સૂત્ર $2 	an ^{-1} x = 	an ^{-1} \left( \frac{2x}{1-x^2} ight)$ નો ઉપયોગ કરીએ છીએ.$x = \frac{1}{3}$ માટે,આપણને મળે છે:$2 	an ^{-1} \left( \frac{1}{3} ight) = 	an ^{-1} \left( \frac{2 	imes \frac{1}{3}}{1 - (\frac{1}{3})^2} ight)$$= 	an ^{-1} \left( \frac{\frac{2}{3}}{1 - \frac{1}{9}} ight)$$= 	an ^{-1} \left( \frac{\frac{2}{3}}{\frac{8}{9}} ight)$$= 	an ^{-1} \left( \frac{2}{3} 	imes \frac{9}{8} ight) = 	an ^{-1} \left( \frac{3}{4} ight)$હવે,આ કિંમતને મૂળ પદાવલિમાં મૂકતા:$2 	an ^{-1} \left( \frac{1}{3} ight) - 	an ^{-1} \left( \frac{3}{4} ight) = 	an ^{-1} \left( \frac{3}{4} ight) - 	an ^{-1} \left( \frac{3}{4} ight) = 0$.

$2 \tan ^{-1}\left(\frac{1}{3}\right)-\tan ^{-1}\left(\frac{3}{4}\right)=$

Similar Questions

જો $y = \cot^{-1} \left( \frac{1 + \sin 5x}{\cos 5x} \right)$ હોય, તો $\frac{dy}{dx}$ નું મૂલ્ય છે

$\tan ^{-1} 2 + \tan ^{-1} 3 = $

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

$\operatorname{Sin}^{-1}(-\cos 2) + \operatorname{Cos}^{-1}(\sin 3) + \operatorname{Tan}^{-1}(\cot 5) = $

$\begin{aligned} & 2 \sin ^{-1} x+\sin ^{-1}\left(2 x \sqrt{1-x^2}\right)+3 \cos ^{-1} x \\ & -\cos ^{-1}\left(4 x^3-3 x\right) \text{ની કિંમત શોધો. }\end{aligned}$

Vedclass Test Series

Exam Paper Generator

Online Exam Module