Math, asked by aadishree7667, 7 months ago

plz solve this math qn""

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Answered by silentloffer

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$tan {}^{ -1 } \sqrt{3} - sec {}^{ - 1} ( - 2) = > \\ \\ = > you \: know \: ... \: \: sec {}^{ - 1} ( - x ) = \pi - sec { }^{ - 1} x \\ \\ \\ = > tan {}^{ - 1} \sqrt{3} - (\pi - sec {}^{ - 1} (2) = \\ \\ = > \frac{\pi}{3} - \pi + \frac{\pi}{3} = \frac{2\pi}{3} - \pi = \\ \\ \\ = > \frac{2\pi - 3\pi}{3} = \frac{ - \pi}{3} \: \: ans..$

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Answered by rishikaar19

$\begin{gathered}tan {}^{ -1 } \sqrt{3} - sec {}^{ - 1} ( - 2) = > \\ \\ = > you \: know \: ... \: \: sec {}^{ - 1} ( - x ) = \pi - sec { }^{ - 1} x \\ \\ \\ = > tan {}^{ - 1} \sqrt{3} - (\pi - sec {}^{ - 1} (2) = \\ \\ = > \frac{\pi}{3} - \pi + \frac{\pi}{3} = \frac{2\pi}{3} - \pi = \\ \\ \\ = > \frac{2\pi - 3\pi}{3} = \frac{ - \pi}{3} \: \: ans..\end{gathered}tan−13−sec−1(−2)=>=>youknow...sec−1(−x)=π−sec−1x=>tan−13−(π−sec−1(2)==>3π−π+3π=32π−π==>32π−3π=3−πans..$

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