Evaluate: tan^{-1}left(frac{1}{4}right) + tan | vedclass

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Question

(A) Using the formula $	an^{-1}x + 	an^{-1}y = 	an^{-1}\left(\frac{x+y}{1-xy}ight)$,we have:$	an^{-1}\left(\frac{1}{4}ight) + 	an^{-1}\left(\

Vedclass · Accepted Answer

$\frac{1}{2}\cos^{-1}\left(\frac{3}{5}\right)$

Vedclass · Answer

(A) Using the formula $	an^{-1}x + 	an^{-1}y = 	an^{-1}\left(\frac{x+y}{1-xy}ight)$,we have:$	an^{-1}\left(\frac{1}{4}ight) + 	an^{-1}\left(\frac{2}{9}ight) = 	an^{-1}\left(\frac{\frac{1}{4} + \frac{2}{9}}{1 - \frac{1}{4} 	imes \frac{2}{9}}ight)$$= 	an^{-1}\left(\frac{\frac{9+8}{36}}{1 - \frac{2}{36}}ight) = 	an^{-1}\left(\frac{17/36}{34/36}ight) = 	an^{-1}\left(\frac{17}{34}ight) = 	an^{-1}\left(\frac{1}{2}ight)$Now,we use the formula $2	an^{-1}x = \cos^{-1}\left(\frac{1-x^2}{1+x^2}ight)$:$	an^{-1}\left(\frac{1}{2}ight) = \frac{1}{2} \left( 2	an^{-1}\left(\frac{1}{2}ight) ight) = \frac{1}{2} \cos^{-1}\left(\frac{1-(1/2)^2}{1+(1/2)^2}ight)$$= \frac{1}{2} \cos^{-1}\left(\frac{1-1/4}{1+1/4}ight) = \frac{1}{2} \cos^{-1}\left(\frac{3/4}{5/4}ight) = \frac{1}{2} \cos^{-1}\left(\frac{3}{5}ight)$Thus,the correct option is $A$.

Evaluate: $\tan^{-1}\left(\frac{1}{4}\right) + \tan^{-1}\left(\frac{2}{9}\right) = $

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