tan ^{-1}left(frac{1}{8}right)+tan ^{-1}left( | vedclass

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

Question

(B) આપણે સૂત્ર $	an ^{-1}(x) + 	an ^{-1}(y) = 	an ^{-1}\left(\frac{x+y}{1-xy}ight)$ નો ઉપયોગ કરીશું.પ્રથમ,$	an ^{-1}\left(\frac{1}{2}ight) + \

Vedclass · Accepted Answer

$\frac{\pi}{4}$

Vedclass · Answer

(B) આપણે સૂત્ર $	an ^{-1}(x) + 	an ^{-1}(y) = 	an ^{-1}\left(\frac{x+y}{1-xy}ight)$ નો ઉપયોગ કરીશું.પ્રથમ,$	an ^{-1}\left(\frac{1}{2}ight) + 	an ^{-1}\left(\frac{1}{5}ight)$ ની ગણતરી કરો:$	an ^{-1}\left(\frac{\frac{1}{2} + \frac{1}{5}}{1 - \frac{1}{2} \cdot \frac{1}{5}}ight) = 	an ^{-1}\left(\frac{\frac{7}{10}}{\frac{9}{10}}ight) = 	an ^{-1}\left(\frac{7}{9}ight)$.હવે,પરિણામમાં $	an ^{-1}\left(\frac{1}{8}ight)$ ઉમેરો:$	an ^{-1}\left(\frac{1}{8}ight) + 	an ^{-1}\left(\frac{7}{9}ight) = 	an ^{-1}\left(\frac{\frac{1}{8} + \frac{7}{9}}{1 - \frac{1}{8} \cdot \frac{7}{9}}ight)$.કૌંસની અંદરના પદનું સાદું રૂપ આપતા:$\frac{\frac{9+56}{72}}{\frac{72-7}{72}} = \frac{65}{65} = 1$.આમ,પદાવલિ $	an ^{-1}(1) = \frac{\pi}{4}$ થાય છે.

$\tan ^{-1}\left(\frac{1}{8}\right)+\tan ^{-1}\left(\frac{1}{2}\right)+\tan ^{-1}\left(\frac{1}{5}\right)$ નું મૂલ્ય શોધો.

Similar Questions

$S = \tan^{-1}\left( \frac{1}{n^2 + n + 1} \right) + \tan^{-1}\left( \frac{1}{n^2 + 3n + 3} \right) + \dots + \tan^{-1}\left( \frac{1}{1 + (n + 19)(n + 20)} \right)$ હોય,તો $\tan S$ ની કિંમત શોધો.

$\cos \left[\cos ^{-1}\left(-\frac{1}{7}\right)+\sin ^{-1}\left(-\frac{1}{7}\right)\right]$ ની કિંમત શોધો.

જો $y = \tan^{-1}\left(\frac{1}{x^2 + x + 1}\right) + \tan^{-1}\left(\frac{1}{x^2 + 3x + 3}\right) + \tan^{-1}\left(\frac{1}{x^2 + 5x + 7}\right) + \dots$ $n$ પદો સુધી હોય,તો $\frac{dy}{dx}$ ની કિંમત શોધો.

$\tan ^{-1} \sqrt{3} - \cot ^{-1}(-\sqrt{3}) = $ . . . . . . .

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module