cot ^{-1}left(frac{1}{2}right)+cot ^{-1}left( | vedclass

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(D) We know that $\cot ^{-1}(x) = 	an ^{-1}\left(\frac{1}{x}ight)$ for $x > 0$. Given expression is $\cot ^{-1}\left(\frac{1}{2}ight)+\cot ^{-1}\

Vedclass · Accepted Answer

$\frac{3 \pi}{4}$

Vedclass · Answer

(D) We know that $\cot ^{-1}(x) = 	an ^{-1}\left(\frac{1}{x}ight)$ for $x > 0$. Given expression is $\cot ^{-1}\left(\frac{1}{2}ight)+\cot ^{-1}\left(\frac{1}{3}ight)$. This can be written as $	an ^{-1}(2) + 	an ^{-1}(3)$. Since $xy = 2 	imes 3 = 6 > 1$,we use the formula $	an ^{-1}(x) + 	an ^{-1}(y) = \pi + 	an ^{-1}\left(\frac{x+y}{1-xy}ight)$. Substituting the values,we get $\pi + 	an ^{-1}\left(\frac{2+3}{1-6}ight) = \pi + 	an ^{-1}\left(\frac{5}{-5}ight)$. This simplifies to $\pi + 	an ^{-1}(-1) = \pi - \frac{\pi}{4} = \frac{3 \pi}{4}$. Thus,the correct option is $D$.

$\cot ^{-1}\left(\frac{1}{2}\right)+\cot ^{-1}\left(\frac{1}{3}\right)=$ . . . . . . .

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