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Inverse Trigonometric Functions — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsInverse Trigonometric Functions2 Marks
Question
Write the value of $\sin^{-1}\Big(\frac{1}{3}\Big)-\cos^{-1}\Big(-\frac{1}{3}\Big).$
✓

Answer

We know that $\sin^{-1}\text{x}+\cos^{-1}\text{x}=\frac{\pi}{2}$ and $\cos^{-1}(-\text{x})=\pi-\cos^{-1}\text{x}.$
$\therefore\ \sin^{-1}\Big(\frac{1}{3}\Big)-\cos^{-1}\Big(-\frac{1}{3}\Big)$
$=\sin^{-1}\Big(\frac{1}{3}\Big)-\Big[\pi-\cos^{-1}\Big(\frac{1}{3}\Big)\Big]$
$=\sin\Big(\frac{1}{3}\Big)-\pi+\cos^{-1}\Big(\frac{1}{3}\Big)$
$=\Big[\sin^{-1}\Big(\frac{1}{3}\Big)+\cos^{-1}\Big(\frac{1}{3}\Big)\Big]-\pi$
$=\frac{-\pi}{2}$ $\Big[\because\ \sin^{-1}\text{x}+\cos^{-1}\text{x}=\frac{\pi}{2}\Big]$
$=-\frac{\pi}{2}$
$\therefore\ \sin^{-1}\Big(\frac{1}{3}\Big)-\cos^{-1}\Big(-\frac{1}{3}\Big)=-\frac{\pi}{2}$

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