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Question 164 of 177

Q.If −1≤x≤1-1 \le x \le 1, then prove that sin⁡−1x+cos⁡−1x=π2\sin^{-1}x + \cos^{-1}x = \dfrac{\pi}{2}

Maharashtra MsbshseMaharashtra HSC (MSBSHSE) Board 2023Subjective· 3mImportance★★★★★
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Let y=cos⁡−1xy=\cos^{-1}x and show π2−y=sin⁡−1x\frac{\pi}2-y=\sin^{-1}x.

Let y=cos⁡−1xy=\cos^{-1}x, so x=cos⁡yx=\cos y, with y∈[0,π]y\in[0,\pi].

Then x=cos⁡y=sin⁡(π2−y)x=\cos y=\sin\left(\dfrac{\pi}{2}-y\right).

Since y∈[0,π]y\in[0,\pi], we have π2−y∈[−π2,π2]\dfrac{\pi}{2}-y\in\left[-\dfrac{\pi}{2},\dfrac{\pi}{2}\right], which is exactly the range of sin⁡−1\sin^{-1}.

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