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INVERSE TRIGNOMETRIC FUNCTIONS — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSINVERSE TRIGNOMETRIC FUNCTIONS3 Marks
Question
$\text{Prove that:} \sin^{-1} \bigg(\frac{4}{5}\bigg) + \sin^{-1} \bigg(\frac{5}{13}\bigg) + \sin^{-1} \bigg(\frac{16}{65}\bigg) = \frac{\pi}{2}$
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Answer

$\text{Getting} \sin^{-1}\Big(\frac{4}{5}\Big) = \tan^{-1} \Big(\frac{4}{3}\Big), \sin^{-1}\Big(\frac{5}{13}\Big) = \tan^{-1} \Big(\frac{5}{12}\Big), \sin^{-1} \Big(\frac{16}{65}\Big) = \tan^{-1} \Big(\frac{16}{63}\Big)$$\text{LHS} = \tan^{-1} \Big(\frac{4}{3}\Big) + \tan^{-1}\Big(\frac{5}{12}\Big) + \tan^{-1} \Big(\frac{16}{63}\Big) = \tan^{-1} \Bigg (\frac{\frac{4}{3} + \frac{5}{12}}{1 - \frac{4}{3}. \frac{5}{12}}\Bigg) + \tan^{-1} \Big(\frac{16}{63}\Big)$
$= \tan^{-1} \bigg(\frac{63}{16}\bigg) + \cot^{-1} \bigg(\frac{63}{16}\bigg)$
$=\frac{\pi}{2}$

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