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If $\sin^{-1}(x-1)+\cos^{-1}(x-3)+\tan^{-1}\left ( \frac{x}{2-x^{2}} \right )=\cos^{-1}k+\pi,$ then the value of k is equal to (Up to one decimal point)
Option: 1
$1$

Option: 2
$-\frac{1}{\sqrt{2}}$

Option: 3
0.7

Option: 4
2

Answers (1)

best_answer

Domains and Ranges of Inverse Trigonometric Functions -

For $\sin ^{-1}x$

Domain $\epsilon \left [ -1, 1 \right ]$

Range $\epsilon \left [ -\frac{\pi }{2}, \frac{\pi }{2} \right ]$

-

Domains and Ranges of Inverse Trigonometric Functions -

For $\cos ^{-1}x$

Domain $\epsilon \left [ -1, 1 \right ]$

Range $\epsilon \left [ 0, \pi \right ]$

-

Domains and Ranges of Inverse Trigonometric Functions -

For $\tan ^{-1}x$

Domain $\epsilon R$

Range $\epsilon\left [ -\frac{\pi }{2},\frac{\pi }{2} \\ \right ]$

-

$\sin ^{-1}(x-1)\Rightarrow -1\leq x-1\leq 1\Rightarrow 0\leq x\leq 2$

$\cos ^{-1}(x-3)\Rightarrow -1\leq x-3\leq 1\Rightarrow 2\leq x\leq 4$

$\tan ^{-1}\left ( \frac{x}{2-x^{2}} \right )\Rightarrow x\: \in \; R, x\neq \sqrt 2 ,$

$\therefore x=2$

$\sin ^{-1}(2-1)+\cos ^{-1}(2-3)+\tan ^{-1}\frac{2}{2-4}=\cos ^{-1}k+\pi$

$\sin ^{-1}1+\cos ^{-1}(-1)+\tan ^{-1}(-1)=\cos ^{-1}k+\pi$

$\frac{\pi}{2}+\pi-\frac{\pi}{4}=\cos ^{-1}k+\pi$

$\Rightarrow \cos ^{-1}k=\frac{\pi}{4}\Rightarrow k=\frac{1}{\sqrt{2}}$

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Sanket Gandhi

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