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

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

Question

(D) We know that $\cot^{-1}(x) = 	an^{-1}(\frac{1}{x})$ for $x > 0$.Therefore,the expression becomes:$	an^{-1}(1) + 	an^{-1}(2) + 	an^{-1}(3)$We k

Vedclass · Accepted Answer

$\pi$

Vedclass · Answer

(D) We know that $\cot^{-1}(x) = 	an^{-1}(\frac{1}{x})$ for $x > 0$.Therefore,the expression becomes:$	an^{-1}(1) + 	an^{-1}(2) + 	an^{-1}(3)$We know that $	an^{-1}(1) = \frac{\pi}{4}$.Now,for $	an^{-1}(2) + 	an^{-1}(3)$,since $2 	imes 3 = 6 > 1$,we use the formula $	an^{-1}(x) + 	an^{-1}(y) = \pi + 	an^{-1}(\frac{x+y}{1-xy})$:$	an^{-1}(2) + 	an^{-1}(3) = \pi + 	an^{-1}(\frac{2+3}{1-2 	imes 3})$$= \pi + 	an^{-1}(\frac{5}{1-6})$$= \pi + 	an^{-1}(\frac{5}{-5})$$= \pi + 	an^{-1}(-1)$$= \pi - 	an^{-1}(1)$$= \pi - \frac{\pi}{4} = \frac{3\pi}{4}$.Adding these results together:$\frac{\pi}{4} + \frac{3\pi}{4} = \frac{4\pi}{4} = \pi$.

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

Similar Questions

If $0 \leq x < \frac{3}{4}$, then the number of values of $x$ satisfying the equation $\operatorname{Tan}^{-1}(2x-1) + \operatorname{Tan}^{-1}(2x) = \operatorname{Tan}^{-1}(4x) - \operatorname{Tan}^{-1}(2x+1)$ is:

The value of $\lim_{x \to 1/\sqrt{2}} \frac{x - \cos(\sin^{-1} x)}{1 - \tan(\sin^{-1} x)}$ is

The possible values of $x$, which satisfy the trigonometric equation $\tan ^{-1}\left(\frac{x-1}{x-2}\right)+\tan ^{-1}\left(\frac{x+1}{x+2}\right)=\frac{\pi}{4}$ are

The derivative of $\tan^{-1} \left( \frac{\sqrt{1+x^2} - 1}{x} \right)$ with respect to $\sin^{-1} \left( \frac{2x}{1+x^2} \right)$ is...

If $\sinh (2 \tanh ^{-1} x) = \frac{11}{60}$,then $x =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module