sin ^{-1}(sin 100) + cos ^{-1}(cos 100) + tan | vedclass

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

Question

(C) We know the principal value ranges: $\sin^{-1}(\sin x) = x - n\pi$ or $(n+1)\pi - x$,$\cos^{-1}(\cos x) = |x - 2n\pi|$,$	an^{-1}(	an x) = x - n\

Vedclass · Accepted Answer

$200 - 63\pi$

Vedclass · Answer

(C) We know the principal value ranges: $\sin^{-1}(\sin x) = x - n\pi$ or $(n+1)\pi - x$,$\cos^{-1}(\cos x) = |x - 2n\pi|$,$	an^{-1}(	an x) = x - n\pi$,and $\cot^{-1}(\cot x) = x - n\pi$.
For $x = 100$,we note that $31\pi \approx 97.389$ and $32\pi \approx 100.53$.
Since $31\pi < 100 < 32\pi$:
$1$. $\sin^{-1}(\sin 100) = 100 - 32\pi$ (as $100$ is in the range where $\sin x$ is negative and decreasing).
$2$. $\cos^{-1}(\cos 100) = 32\pi - 100$ (since $100$ is in the interval $(31\pi, 32\pi)$).
$3$. $	an^{-1}(	an 100) = 100 - 32\pi$ (since $100$ is in the interval $(31\pi, 32\pi)$).
$4$. $\cot^{-1}(\cot 100) = 100 - 31\pi$ (since $100$ is in the interval $(31\pi, 32\pi)$).
Summing these: $(100 - 32\pi) + (32\pi - 100) + (100 - 32\pi) + (100 - 31\pi) = 200 - 63\pi$.

$\sin ^{-1}(\sin 100) + \cos ^{-1}(\cos 100) + \tan ^{-1}(\tan 100) + \cot ^{-1}(\cot 100)$ equals to:

Similar Questions

If $x, y, z$ are in $A.P.$ and $\tan^{-1}x, \tan^{-1}y, \tan^{-1}z$ are also in $A.P.$,then

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

$\lim _{n \rightarrow \infty} \sum_{r=1}^n \cot ^{-1}\left(r^2+\frac{3}{4}\right)=$

If $p$ and $q$ are roots of $6x^2 + 10x + 1 = 0$,then the value of $[\tan^{-1} p + \tan^{-1} q]$ is: {where $[x]$ denotes the greatest integer less than or equal to $x$}

Considering only the principal values of the inverse trigonometric functions,the set $\{x \geq 0 : \tan^{-1}(2x) + \tan^{-1}(3x) = \frac{\pi}{4}\}$

Vedclass Test Series

Exam Paper Generator

Online Exam Module