The value of 2 tan^{-1} frac{1}{2} + tan^{-1} | vedclass

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(C) We use the formula $2 	an^{-1} x = 	an^{-1} \left( \frac{2x}{1-x^2} ight)$ for $|x| < 1$. Applying this to $2 	an^{-1} \frac{1}{2}$: $2 	an^

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$	an^{-1} \left( \frac{31}{17} ight)$

Vedclass · Answer

(C) We use the formula $2 	an^{-1} x = 	an^{-1} \left( \frac{2x}{1-x^2} ight)$ for $|x| Applying this to $2 	an^{-1} \frac{1}{2}$: $2 	an^{-1} \frac{1}{2} = 	an^{-1} \left( \frac{2(1/2)}{1-(1/2)^2} ight) = 	an^{-1} \left( \frac{1}{1-1/4} ight) = 	an^{-1} \left( \frac{1}{3/4} ight) = 	an^{-1} \left( \frac{4}{3} ight)$. Now,add $	an^{-1} \frac{1}{7}$ using the formula $	an^{-1} x + 	an^{-1} y = 	an^{-1} \left( \frac{x+y}{1-xy} ight)$: $	an^{-1} \left( \frac{4}{3} ight) + 	an^{-1} \left( \frac{1}{7} ight) = 	an^{-1} \left( \frac{4/3 + 1/7}{1 - (4/3)(1/7)} ight)$. Calculate the numerator: $\frac{4}{3} + \frac{1}{7} = \frac{28+3}{21} = \frac{31}{21}$. Calculate the denominator: $1 - \frac{4}{21} = \frac{21-4}{21} = \frac{17}{21}$. Thus,the expression becomes $	an^{-1} \left( \frac{31/21}{17/21} ight) = 	an^{-1} \left( \frac{31}{17} ight)$.

The value of $2 \tan^{-1} \frac{1}{2} + \tan^{-1} \frac{1}{7}$ is:

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