operatorname{Tan}^{-1} frac{3}{5} + operatorn | vedclass

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Question

(A) हम सूत्र $\operatorname{Tan}^{-1} x + \operatorname{Tan}^{-1} y = \operatorname{Tan}^{-1} \left( \frac{x+y}{1-xy} \right)$ का उपयोग करते हैं।सबसे

Vedclass · Accepted Answer

$\operatorname{Tan}^{-1} \frac{9}{10}$

Vedclass · Answer

(A) हम सूत्र $\operatorname{Tan}^{-1} x + \operatorname{Tan}^{-1} y = \operatorname{Tan}^{-1} \left( \frac{x+y}{1-xy} ight)$ का उपयोग करते हैं।सबसे पहले, $\operatorname{Tan}^{-1} \frac{3}{5} + \operatorname{Tan}^{-1} \frac{6}{41}$ की गणना करें:$= \operatorname{Tan}^{-1} \left( \frac{\frac{3}{5} + \frac{6}{41}}{1 - \frac{3}{5} 	imes \frac{6}{41}} ight) = \operatorname{Tan}^{-1} \left( \frac{\frac{123+30}{205}}{\frac{205-18}{205}} ight) = \operatorname{Tan}^{-1} \left( \frac{153}{187} ight) = \operatorname{Tan}^{-1} \left( \frac{9}{11} ight)$.अब, तीसरा पद जोड़ें: $\operatorname{Tan}^{-1} \frac{9}{11} + \operatorname{Tan}^{-1} \frac{9}{191}$:$= \operatorname{Tan}^{-1} \left( \frac{\frac{9}{11} + \frac{9}{191}}{1 - \frac{9}{11} 	imes \frac{9}{191}} ight) = \operatorname{Tan}^{-1} \left( \frac{\frac{1719+99}{2101}}{\frac{2101-81}{2101}} ight) = \operatorname{Tan}^{-1} \left( \frac{1818}{2020} ight) = \operatorname{Tan}^{-1} \left( \frac{9}{10} ight)$.

$\operatorname{Tan}^{-1} \frac{3}{5} + \operatorname{Tan}^{-1} \frac{6}{41} + \operatorname{Tan}^{-1} \frac{9}{191} = $

Similar Questions

मूल्यांकन करें: $\tan ^{ - 1}\left(\frac{{{c_1}x - y}}{{{c_1}y + x}}\right) + \tan ^{ - 1}\left(\frac{{{c_2} - {c_1}}}{{1 + {c_2}{c_1}}}\right) + \tan ^{ - 1}\left(\frac{{{c_3} - {c_2}}}{{1 + {c_3}{c_2}}}\right) + ... + \tan ^{ - 1}\left(\frac{1}{{{c_n}}}\right)$

यदि $\sin ^{-1}\left(\frac{3}{x}\right)+\sin ^{-1}\left(\frac{4}{x}\right)=\frac{\pi}{2}$ है, तो $x$ का मान ज्ञात कीजिए।

यदि $y = \operatorname{cosec}^{-1}\left[\frac{\sqrt{x}+1}{\sqrt{x}-1}\right] + \cos^{-1}\left[\frac{\sqrt{x}-1}{\sqrt{x}+1}\right]$ है,तो $\frac{dy}{dx} = $

$\sin ^{-1}\left(-\frac{1}{\sqrt{2}}\right)+\cos ^{-1}\left(-\frac{1}{2}\right)-\cot ^{-1}\left(-\frac{1}{\sqrt{3}}\right)+\tan ^{-1}(\sqrt{3})$ का मान ज्ञात कीजिए।

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

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