cos left[ {{tan }^{ - 1}}frac{1}{3} + {{tan } | vedclass

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(A) We use the formula $	an^{-1} x + 	an^{-1} y = 	an^{-1} \left( \frac{x+y}{1-xy} ight)$ for $xy < 1$.Here,$x = \frac{1}{3}$ and $y = \frac{1}{2

Vedclass · Accepted Answer

$\frac{1}{{\sqrt 2 }}$

Vedclass · Answer

(A) We use the formula $	an^{-1} x + 	an^{-1} y = 	an^{-1} \left( \frac{x+y}{1-xy} ight)$ for $xy Here,$x = \frac{1}{3}$ and $y = \frac{1}{2}$.Since $\frac{1}{3} 	imes \frac{1}{2} = \frac{1}{6} $	an^{-1} \frac{1}{3} + 	an^{-1} \frac{1}{2} = 	an^{-1} \left( \frac{\frac{1}{3} + \frac{1}{2}}{1 - \frac{1}{3} 	imes \frac{1}{2}} ight) = 	an^{-1} \left( \frac{\frac{5}{6}}{1 - \frac{1}{6}} ight) = 	an^{-1} \left( \frac{\frac{5}{6}}{\frac{5}{6}} ight) = 	an^{-1}(1) = \frac{\pi}{4}$.Therefore,$\cos \left[ 	an^{-1} \frac{1}{3} + 	an^{-1} \frac{1}{2} ight] = \cos \left( \frac{\pi}{4} ight) = \frac{1}{\sqrt{2}}$.

$\cos \left[ {{\tan }^{ - 1}}\frac{1}{3} + {{\tan }^{ - 1}}\frac{1}{2} \right] = $

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