4 tan^{-1} frac{1}{5} - tan^{-1} frac{1}{70} | vedclass

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

Question

(C) આપણે સૂત્ર $2 	an^{-1} x = 	an^{-1} \left( \frac{2x}{1-x^2} ight)$ નો ઉપયોગ કરીશું.પ્રથમ,$4 	an^{-1} \frac{1}{5} = 2 \left( 2 	an^{-1} \frac

Vedclass · Accepted Answer

$\frac{\pi}{4}$

Vedclass · Answer

(C) આપણે સૂત્ર $2 	an^{-1} x = 	an^{-1} \left( \frac{2x}{1-x^2} ight)$ નો ઉપયોગ કરીશું.પ્રથમ,$4 	an^{-1} \frac{1}{5} = 2 \left( 2 	an^{-1} \frac{1}{5} ight) = 2 	an^{-1} \left( \frac{2/5}{1 - 1/25} ight) = 2 	an^{-1} \left( \frac{2/5}{24/25} ight) = 2 	an^{-1} \left( \frac{5}{12} ight)$.હવે,$2 	an^{-1} \frac{5}{12} = 	an^{-1} \left( \frac{2(5/12)}{1 - (5/12)^2} ight) = 	an^{-1} \left( \frac{5/6}{1 - 25/144} ight) = 	an^{-1} \left( \frac{5/6}{119/144} ight) = 	an^{-1} \left( \frac{5}{6} 	imes \frac{144}{119} ight) = 	an^{-1} \left( \frac{120}{119} ight)$.આગળ,આપણે $	an^{-1} \frac{1}{99} - 	an^{-1} \frac{1}{70} = 	an^{-1} \left( \frac{1/99 - 1/70}{1 + (1/99)(1/70)} ight) = 	an^{-1} \left( \frac{(70-99)/6930}{(6930+1)/6930} ight) = 	an^{-1} \left( \frac{-29}{6931} ight) = -	an^{-1} \left( \frac{1}{239} ight)$ ની ગણતરી કરીએ.છેલ્લે,$	an^{-1} \frac{120}{119} - 	an^{-1} \frac{1}{239} = 	an^{-1} \left( \frac{120/119 - 1/239}{1 + (120/119)(1/239)} ight) = 	an^{-1} \left( \frac{(28680 - 119) / 28441}{(28441 + 120) / 28441} ight) = 	an^{-1} \left( \frac{28561}{28561} ight) = 	an^{-1}(1) = \frac{\pi}{4}$.

$4 \tan^{-1} \frac{1}{5} - \tan^{-1} \frac{1}{70} + \tan^{-1} \frac{1}{99} = $

Similar Questions

$\cos \left(\cos ^{-1} \frac{1}{3}+\cos ^{-1} \frac{1}{5}\right)+\cos \left(\sin ^{-1} \frac{1}{3}+\sin ^{-1} \frac{1}{5}\right) =$ . . . . . . .

કિંમત શોધો: ${\tan ^{ - 1}}1 + {\tan ^{ - 1}}2 + {\tan ^{ - 1}}3$

$\tan ^{-1}\left(\frac{x}{y}\right)-\tan ^{-1}\left(\frac{x-y}{x+y}\right)$ (જ્યાં,$x, y>0$) નું મૂલ્ય શું છે?

જો $\sin^{-1} x + \sin^{-1} y = \pi/2$ હોય, તો $x^2$ કોના બરાબર છે?

$x$ ના મૂલ્યોનો ગણ શોધો જેથી $\tan ^{-1}\left(\frac{x}{x-2}\right)-\tan ^{-1}\left(\frac{x}{2 x-1}\right)=\tan ^{-1}\left(\frac{2}{3}\right)$ થાય.

Vedclass Test Series

Exam Paper Generator

Online Exam Module