Evaluate: {tan ^{ - 1}}x + {cot ^{ - 1}}(x + | vedclass

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

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${	an ^{ - 1}}({x^2} + x + 1)$

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(D) We know that ${\cot ^{ - 1}}(y) = {	an ^{ - 1}}\left( \frac{1}{y} ight)$ for $y > 0$. Given expression is ${	an ^{ - 1}}x + {\cot ^{ - 1}}(x + 1)$. Using the property,we can write this as: ${	an ^{ - 1}}x + {	an ^{ - 1}}\left( \frac{1}{x + 1} ight)$. Now,apply the formula ${	an ^{ - 1}}A + {	an ^{ - 1}}B = {	an ^{ - 1}}\left( \frac{A + B}{1 - AB} ight)$: $= {	an ^{ - 1}}\left( \frac{x + \frac{1}{x + 1}}{1 - x \cdot \frac{1}{x + 1}} ight)$ $= {	an ^{ - 1}}\left( \frac{\frac{x(x + 1) + 1}{x + 1}}{\frac{x + 1 - x}{x + 1}} ight)$ $= {	an ^{ - 1}}\left( \frac{x^2 + x + 1}{1} ight)$ $= {	an ^{ - 1}}({x^2} + x + 1)$.

Evaluate: ${\tan ^{ - 1}}x + {\cot ^{ - 1}}(x + 1)$

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