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

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(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:

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