If theta = sin^{-1}x + cos^{-1}x - tan^{-1}x | vedclass

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Question

(B) We are given the expression $	heta = \sin^{-1}x + \cos^{-1}x - 	an^{-1}x$ for $x \ge 0$. Using the identity $\sin^{-1}x + \cos^{-1}x = \frac{\pi

Vedclass · Accepted Answer

$0 < 	heta \le \frac{\pi}{2}$

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(B) We are given the expression $	heta = \sin^{-1}x + \cos^{-1}x - 	an^{-1}x$ for $x \ge 0$. Using the identity $\sin^{-1}x + \cos^{-1}x = \frac{\pi}{2}$ for $x \in [-1, 1]$,the expression simplifies to: $	heta = \frac{\pi}{2} - 	an^{-1}x$. Since $x \ge 0$,the range of $	an^{-1}x$ is $0 \le 	an^{-1}x Multiplying by $-1$,we get $-\frac{\pi}{2} Adding $\frac{\pi}{2}$ to all parts,we get $\frac{\pi}{2} - \frac{\pi}{2} Thus,$0 Therefore,the correct interval is $0 < 	heta \le \frac{\pi}{2}$.

If $\theta = \sin^{-1}x + \cos^{-1}x - \tan^{-1}x$ for $x \ge 0$,then the smallest interval in which $\theta$ lies is given by

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