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Inverse Trigonometric Functions — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsInverse Trigonometric Functions5 Marks
Question
Evaluate the following:
$\sin\Big(\sec^{-1}\frac{17}{8}\Big)$
✓

Answer

$\sin\Big(\sec^{-1}\frac{17}{8}\Big)$
$=\sin\Big(\cos^{-1}\frac{8}{17}\Big)$
$=\sin\Bigg[\sin^{-1}\sqrt{1-\Big(\frac{8}{17}\Big)^2}\Bigg]$ $\Big[{\therefore\ \cos^{-1}}\text{x}=\sin^{-1}\sqrt{1-\text{x}^2}\Big]$
$=\sin\Bigg[\sin^{-1}\Bigg(\sqrt{1-\frac{64}{289}}\Bigg)\Bigg]$
$=\sin\Bigg[\sin^{-1}\Bigg(\sqrt{\frac{225}{289}}\Bigg)\Bigg]$
$=\sin\Big[\sin^{-1}\frac{15}{17}\Big]$
$=\frac{15}{17}$

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