If sin^{-1}(x - 2) + cos^{-1}(x) + tan^{-1}(x | vedclass

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(B) The domain of $\sin^{-1}(x-2)$ is $[1, 3]$, $\cos^{-1}(x)$ is $[-1, 1]$, $	an^{-1}(x+2)$ is $(-\infty, \infty)$, and $\cot^{-1}(x+4)$ is $(-\inft

Vedclass · Accepted Answer

$\frac{31}{32}$

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(B) The domain of $\sin^{-1}(x-2)$ is $[1, 3]$, $\cos^{-1}(x)$ is $[-1, 1]$, $	an^{-1}(x+2)$ is $(-\infty, \infty)$, and $\cot^{-1}(x+4)$ is $(-\infty, \infty)$.The intersection of these domains is $\{1\}$.Substituting $x = 1$ into the equation:$\sin^{-1}(1-2) + \cos^{-1}(1) + 	an^{-1}(1+2) + \cot^{-1}(1+4) = \sec^{-1}(\sqrt{k}) - \frac{\pi}{2}$$-\frac{\pi}{2} + 0 + 	an^{-1}(3) + \cot^{-1}(5) = \sec^{-1}(\sqrt{k}) - \frac{\pi}{2}$$	an^{-1}(3) + \cot^{-1}(5) = \sec^{-1}(\sqrt{k})$Using $	an^{-1}(3) = \cot^{-1}(1/3)$, we have $\cot^{-1}(1/3) + \cot^{-1}(5) = \sec^{-1}(\sqrt{k})$.Let $\alpha = \cot^{-1}(1/3)$ and $\beta = \cot^{-1}(5)$. Then $\cot(\alpha + \beta) = \frac{\cot \alpha \cot \beta - 1}{\cot \alpha + \cot \beta} = \frac{(1/3)(5) - 1}{1/3 + 5} = \frac{2/3}{16/3} = \frac{1}{8}$.So, $\alpha + \beta = \cot^{-1}(1/8) = 	an^{-1}(8) = \sec^{-1}(\sqrt{1+8^2}) = \sec^{-1}(\sqrt{65})$.Thus, $\sqrt{k} = \sqrt{65}$, so $k = 65$.We need to find $\cos(2 \csc^{-1}\sqrt{65-1}) = \cos(2 \csc^{-1}(8))$.Let $	heta = \csc^{-1}(8)$, so $\csc 	heta = 8$, which means $\sin 	heta = 1/8$.Then $\cos(2	heta) = 1 - 2\sin^2 	heta = 1 - 2(1/8)^2 = 1 - 2/64 = 1 - 1/32 = \frac{31}{32}$.

If $\sin^{-1}(x - 2) + \cos^{-1}(x) + \tan^{-1}(x + 2) + \cot^{-1}(x + 4) = \sec^{-1}(\sqrt{k}) - \frac{\pi}{2}$, then $\cos(2 \csc^{-1}\sqrt{k - 1}) = \dots$

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