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

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(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

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જો $4 \sin ^{-1} x + 6 \cos ^{-1} x = 3 \pi$ હોય,તો $x = \ldots$.

${\tan ^{ - 1}}\left[ {\frac{{\sqrt {1 + {x^2}} + \sqrt {1 - {x^2}} }}{{\sqrt {1 + {x^2}} - \sqrt {1 - {x^2}} }}} \right]$,જ્યાં $|x| < 1$ અને $x \ne 0$ હોય,તેની કિંમત શું થાય?

જો $y = \sin^{-1}\left(\frac{x^2 - 1}{x^2 + 1}\right) + \sec^{-1}\left(\frac{x^2 + 1}{x^2 - 1}\right)$,$|x| > 1$ હોય,તો $\frac{dy}{dx}$ ની કિંમત શોધો:

જો $4\sin^{-1}x + \cos^{-1}x = \pi$ હોય,તો $x$ ની કિંમત શોધો.

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