cos ^{-1}left{cot left(sum_{i=1}^3 cot ^{-1} | vedclass

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

Question

(D) We need to evaluate the sum $S = \sum_{i=1}^3 \cot ^{-1} i = \cot ^{-1} 1 + \cot ^{-1} 2 + \cot ^{-1} 3$. Using the identity $\cot ^{-1} x = 	an

Vedclass · Accepted Answer

$\frac{\pi}{2}$

Vedclass · Answer

(D) We need to evaluate the sum $S = \sum_{i=1}^3 \cot ^{-1} i = \cot ^{-1} 1 + \cot ^{-1} 2 + \cot ^{-1} 3$. Using the identity $\cot ^{-1} x = 	an ^{-1} \frac{1}{x}$ for $x > 0$,we have $S = 	an ^{-1} 1 + 	an ^{-1} \frac{1}{2} + 	an ^{-1} \frac{1}{3}$. Since $	an ^{-1} 1 = \frac{\pi}{4}$,we calculate $	an ^{-1} \frac{1}{2} + 	an ^{-1} \frac{1}{3} = 	an ^{-1} \left( \frac{1/2 + 1/3}{1 - (1/2)(1/3)} ight) = 	an ^{-1} \left( \frac{5/6}{5/6} ight) = 	an ^{-1} 1 = \frac{\pi}{4}$. Thus,$S = \frac{\pi}{4} + \frac{\pi}{4} = \frac{\pi}{2}$. Now,we substitute this into the expression: $\cos ^{-1} \{ \cot (S) \} = \cos ^{-1} \{ \cot (\frac{\pi}{2}) \}$. Since $\cot (\frac{\pi}{2}) = 0$,the expression becomes $\cos ^{-1} (0)$. We know that $\cos ^{-1} (0) = \frac{\pi}{2}$. Therefore,the correct option is $D$.

$\cos ^{-1}\left\{\cot \left(\sum_{i=1}^3 \cot ^{-1} i\right)\right\}=$ . . . . . . .

Similar Questions

If $x>0, y>0, z>0, xy+yz+zx < 1$ and if $\tan^{-1} x + \tan^{-1} y + \tan^{-1} z = \pi$, then $x+y+z$ equals to

Find the value of $\tan \frac{1}{2} \left[ \sin^{-1} \frac{2x}{1+x^2} + \cos^{-1} \frac{1-y^2}{1+y^2} \right]$,where $|x| <1, y>0$ and $xy < 1$.

If $\theta = \cot^{-1}(7) + \cot^{-1}(8) + \cot^{-1}(18)$,then $\cot \theta$ is equal to

If $x * y = x^{2} + y^{3}$ and $(x * 1) * 1 = x * (1 * 1)$,then the value of $2 \sin^{-1}\left(\frac{x^{4} + x^{2} - 2}{x^{4} + x^{2} + 2}\right)$ is

$\cot ^{ - 1}\left[ \frac{\sqrt {1 - \sin x} + \sqrt {1 + \sin x}}{\sqrt {1 - \sin x} - \sqrt {1 + \sin x}} \right] = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module