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Inverse Trigonometric Functions — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsInverse Trigonometric Functions3 Marks
Question
Write the following in the simplest form:
$\sin^{-1}\Big\{\frac{\text{x}+\sqrt{1-\text{x}^2}}{\sqrt{2}}\Big\},-1<\text{x}<1$
✓

Answer

Let, $\text{x}=\text{a}\sin\theta$
Now,
$\sin^{-1}\Big\{\frac{\text{x}+\sqrt{1-\text{x}^2}}{\sqrt{2}}\Big\}=\sin^{-1}\Big\{\frac{\sin\theta+\sqrt{1-\sin^2\theta}}{\sqrt{2}}\Big\}$
$=\sin^{-1}\Big\{\frac{\sin\theta+\cos\theta}{\sqrt2}\Big\}$
$=\sin^{-1}\Big\{\frac{1}{\sqrt2}\sin\theta=\frac{1}{\sqrt2}\cos\theta\Big\}$
$=\sin^{-1}\Big\{\cos\frac{\pi}{4}\sin\theta+\sin\frac{\pi}{4}\cos\theta\Big\}$
$=\sin^{-1}\Big\{\sin\Big(\theta+\frac{\pi}{4}\Big)\Big\}$
$=\theta+\frac{\pi}{4}$
$=\frac{\pi}{4}=\sin^{-1}\text{x}$
$\therefore\ \sin^{-1}\bigg\{\frac{\text{x}+\sqrt{1-\text{x}^2}}{\sqrt{2}}\bigg\}=\cos^{-1}\text{x}+\frac{\pi}{4}$

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