tan ^{-1} 2 + cot ^{-1}(-3) + cot ^{-1} frac{ | vedclass

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(D) We use the properties $\cot ^{-1}(-x) = \pi - \cot ^{-1} x$ and $	an ^{-1}(-x) = -	an ^{-1} x$ and $\cot ^{-1} x = 	an ^{-1} \frac{1}{x}$ for $

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$\frac{3 \pi}{2}$

Vedclass · Answer

(D) We use the properties $\cot ^{-1}(-x) = \pi - \cot ^{-1} x$ and $	an ^{-1}(-x) = -	an ^{-1} x$ and $\cot ^{-1} x = 	an ^{-1} \frac{1}{x}$ for $x > 0$. Given expression: $E = 	an ^{-1} 2 + \cot ^{-1}(-3) + \cot ^{-1} \frac{1}{3} + 	an ^{-1}\left(-\frac{1}{2}ight)$ Using the properties: $E = 	an ^{-1} 2 + (\pi - \cot ^{-1} 3) + 	an ^{-1} 3 - 	an ^{-1} \frac{1}{2}$ Since $\cot ^{-1} 3 = 	an ^{-1} \frac{1}{3}$,we have: $E = \pi + (	an ^{-1} 2 - 	an ^{-1} \frac{1}{2}) + (	an ^{-1} 3 - 	an ^{-1} \frac{1}{3})$ Using $	an ^{-1} x - 	an ^{-1} y = 	an ^{-1} \frac{x-y}{1+xy}$: $E = \pi + 	an ^{-1} \left( \frac{2 - 1/2}{1 + 2(1/2)} ight) + 	an ^{-1} \left( \frac{3 - 1/3}{1 + 3(1/3)} ight)$ $E = \pi + 	an ^{-1} \left( \frac{3/2}{2} ight) + 	an ^{-1} \left( \frac{8/3}{2} ight)$ $E = \pi + 	an ^{-1} \frac{3}{4} + 	an ^{-1} \frac{4}{3}$ Since $	an ^{-1} \frac{4}{3} = \cot ^{-1} \frac{3}{4}$: $E = \pi + (	an ^{-1} \frac{3}{4} + \cot ^{-1} \frac{3}{4})$ Using $	an ^{-1} x + \cot ^{-1} x = \frac{\pi}{2}$: $E = \pi + \frac{\pi}{2} = \frac{3 \pi}{2}$

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

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