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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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^

Vedclass · Accepted Answer

$	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:

Similar Questions

Suppose $\tan ^{-1} y = \tan ^{-1} x + \tan ^{-1} \left( \frac{2x}{1 - x^2} \right)$,where $|x| < \frac{1}{\sqrt{3}}$,then one of the values of $y$ is

If $2 \tan^{-1}(\cos x) = \tan^{-1}(\csc^2 x)$,then $x =$

$\sin \left(\cos ^{-1}\left(-\frac{1}{7}\right)+\sin ^{-1}\left(-\frac{1}{7}\right)\right) = $ . . . . . . .

$\cos ^{-1}\left(\frac{-1}{2}\right)-2 \sin ^{-1}\left(\frac{1}{2}\right)+3 \cos ^{-1}\left(\frac{-1}{\sqrt{2}}\right)-4 \tan ^{-1}(-1)$ equals

$\cos \left[\cos ^{-1}\left(-\frac{1}{7}\right)+\sin ^{-1}\left(-\frac{1}{7}\right)\right]$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module