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} = $

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