tan left[ 2tan^{-1}left( frac{1}{5} right) - | vedclass

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

Question

(D) આપણે સૂત્ર $2	an^{-1}(x) = 	an^{-1}\left( \frac{2x}{1-x^2} ight)$ નો ઉપયોગ કરીશું.પ્રથમ,$2	an^{-1}\left( \frac{1}{5} ight)$ ની ગણતરી કરો:$2

Vedclass · Accepted Answer

$-\frac{7}{17}$

Vedclass · Answer

(D) આપણે સૂત્ર $2	an^{-1}(x) = 	an^{-1}\left( \frac{2x}{1-x^2} ight)$ નો ઉપયોગ કરીશું.પ્રથમ,$2	an^{-1}\left( \frac{1}{5} ight)$ ની ગણતરી કરો:$2	an^{-1}\left( \frac{1}{5} ight) = 	an^{-1}\left( \frac{2(1/5)}{1-(1/5)^2} ight) = 	an^{-1}\left( \frac{2/5}{1-1/25} ight) = 	an^{-1}\left( \frac{2/5}{24/25} ight) = 	an^{-1}\left( \frac{2}{5} 	imes \frac{25}{24} ight) = 	an^{-1}\left( \frac{5}{12} ight)$.હવે,આ કિંમતને પદાવલિમાં મૂકો:$	an \left[ 	an^{-1}\left( \frac{5}{12} ight) - \frac{\pi}{4} ight] = 	an \left[ 	an^{-1}\left( \frac{5}{12} ight) - 	an^{-1}(1) ight]$.સૂત્ર $	an^{-1}(x) - 	an^{-1}(y) = 	an^{-1}\left( \frac{x-y}{1+xy} ight)$ નો ઉપયોગ કરતા:$	an \left[ 	an^{-1}\left( \frac{5/12 - 1}{1 + (5/12)(1)} ight) ight] = \frac{5/12 - 1}{1 + 5/12} = \frac{-7/12}{17/12} = -\frac{7}{17}$.

$\tan \left[ 2\tan^{-1}\left( \frac{1}{5} \right) - \frac{\pi}{4} \right] = $

Similar Questions

જો $y = \sin^{-1} \left( \frac{5x + 12\sqrt{1-x^2}}{13} \right)$ હોય,તો $\frac{dy}{dx} = $

જો ${\sin ^{ - 1}}\left( {\frac{{2a}}{{1 + {a^2}}}} \right) + {\sin ^{ - 1}}\left( {\frac{{2b}}{{1 + {b^2}}}} \right) = 2{\tan ^{ - 1}}x,$ હોય,તો $x = $

જો $\sec^{-1} x = \csc^{-1} y$ હોય,તો $\cos^{-1} \frac{1}{x} + \cos^{-1} \frac{1}{y} = $

જો $\cos ^{-1} \sqrt{p}+\cos ^{-1} \sqrt{1-p}+\cos ^{-1} \sqrt{1-q}=\frac{3 \pi}{4}$ હોય,તો $q$ ની કિંમત શોધો.

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module