tan^{-1}left(frac{1}{11}right) + tan^{-1}left | vedclass

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(D) We use the formula $	an^{-1}(x) + 	an^{-1}(y) = 	an^{-1}\left(\frac{x+y}{1-xy}ight)$ for $xy < 1$. Given expression: $	an^{-1}\left(\frac{1}

Vedclass · Accepted Answer

None of these

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(D) We use the formula $	an^{-1}(x) + 	an^{-1}(y) = 	an^{-1}\left(\frac{x+y}{1-xy}ight)$ for $xy Given expression: $	an^{-1}\left(\frac{1}{11}ight) + 	an^{-1}\left(\frac{2}{12}ight)$. Here $x = \frac{1}{11}$ and $y = \frac{2}{12} = \frac{1}{6}$. $	an^{-1}\left(\frac{\frac{1}{11} + \frac{1}{6}}{1 - \frac{1}{11} 	imes \frac{1}{6}}ight) = 	an^{-1}\left(\frac{\frac{6+11}{66}}{1 - \frac{1}{66}}ight)$. $= 	an^{-1}\left(\frac{\frac{17}{66}}{\frac{65}{66}}ight) = 	an^{-1}\left(\frac{17}{65}ight)$. Since $	an^{-1}\left(\frac{17}{65}ight)$ is not among the options,the correct choice is $(d)$.

$\tan^{-1}\left(\frac{1}{11}\right) + \tan^{-1}\left(\frac{2}{12}\right) = $

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