Evaluate: {tan ^{ - 1}}1 + {tan ^{ - 1}}2 + { | vedclass

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(D) We use the formula for ${	an ^{ - 1}}x + {	an ^{ - 1}}y$ when $xy > 1$: ${	an ^{ - 1}}x + {	an ^{ - 1}}y = \pi + {	an ^{ - 1}}\left( {\frac{{

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None of these

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(D) We use the formula for ${	an ^{ - 1}}x + {	an ^{ - 1}}y$ when $xy > 1$: ${	an ^{ - 1}}x + {	an ^{ - 1}}y = \pi + {	an ^{ - 1}}\left( {\frac{{x + y}}{{1 - xy}}} ight)$. First,calculate ${	an ^{ - 1}}2 + {	an ^{ - 1}}3$: Since $2 	imes 3 = 6 > 1$,we have: ${	an ^{ - 1}}2 + {	an ^{ - 1}}3 = \pi + {	an ^{ - 1}}\left( {\frac{{2 + 3}}{{1 - 2 	imes 3}}} ight) = \pi + {	an ^{ - 1}}\left( {\frac{5}{{1 - 6}}} ight) = \pi + {	an ^{ - 1}}\left( {\frac{5}{{ - 5}}} ight) = \pi + {	an ^{ - 1}}( - 1) = \pi - \frac{\pi }{4} = \frac{{3\pi }}{4}$. Now,add ${	an ^{ - 1}}1$: ${	an ^{ - 1}}1 + (\frac{{3\pi }}{4}) = \frac{\pi }{4} + \frac{{3\pi }}{4} = \frac{{4\pi }}{4} = \pi$. Since $\pi$ is not among the options $A, B, C$,the correct answer is $D$.

Evaluate: ${\tan ^{ - 1}}1 + {\tan ^{ - 1}}2 + {\tan ^{ - 1}}3$

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