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

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(B) આપણે જાણીએ છીએ કે $2 	an ^{-1}(x) = 	an ^{-1} \left( \frac{2x}{1-x^2} ight)$.તેથી,$2 	an ^{-1} \left( \frac{1}{5} ight) = 	an ^{-1} \left(

Vedclass · Accepted Answer

$-\frac{7}{17}$

Vedclass · Answer

(B) આપણે જાણીએ છીએ કે $2 	an ^{-1}(x) = 	an ^{-1} \left( \frac{2x}{1-x^2} ight)$.તેથી,$2 	an ^{-1} \left( \frac{1}{5} ight) = 	an ^{-1} \left( \frac{2 	imes \frac{1}{5}}{1 - (\frac{1}{5})^2} ight) = 	an ^{-1} \left( \frac{\frac{2}{5}}{1 - \frac{1}{25}} ight) = 	an ^{-1} \left( \frac{\frac{2}{5}}{\frac{24}{25}} ight) = 	an ^{-1} \left( \frac{2}{5} 	imes \frac{25}{24} ight) = 	an ^{-1} \left( \frac{5}{12} ight)$.હવે,પદાવલિ $	an \left[ 	an ^{-1} \left( \frac{5}{12} ight) - 	an ^{-1} (1) ight]$ બને છે.સૂત્ર $	an ^{-1}(x) - 	an ^{-1}(y) = 	an ^{-1} \left( \frac{x-y}{1+xy} ight)$ નો ઉપયોગ કરતા:$	an ^{-1} \left( \frac{\frac{5}{12} - 1}{1 + \frac{5}{12} 	imes 1} ight) = 	an ^{-1} \left( \frac{-\frac{7}{12}}{\frac{17}{12}} ight) = 	an ^{-1} \left( -\frac{7}{17} ight)$.અંતે,$	an \left[ 	an ^{-1} \left( -\frac{7}{17} ight) ight] = -\frac{7}{17}$.

$\tan \left[2 \tan ^{-1}\left(\frac{1}{5}\right)-\frac{\pi}{4}\right]$ નું મૂલ્ય શોધો.

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સાબિત કરો કે $\tan ^{-1} x+\tan ^{-1} \frac{2 x}{1-x^{2}}=\tan ^{-1}\left(\frac{3 x-x^{3}}{1-3 x^{2}}\right)$,જ્યાં $|x| < \frac{1}{\sqrt{3}}$.

જો $\sin \left(\sin ^{-1} \frac{1}{5}+\cos ^{-1} x\right)=1$ હોય,તો $x$ ની કિંમત શોધો.

કિંમત શોધો: $\tan^{-1} \left( \frac{1 - x^2}{2x} \right) + \cos^{-1} \left( \frac{1 - x^2}{1 + x^2} \right)$

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