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What is the solution of cot(sin^-1x)?(a) \(\frac{\sqrt{1-x^2}}{x}\)(b) x(c) \(\sqrt{1-x^2}\)(d) \(\sqrt{1+x^2}\)This question was posed to me in semester exam.Question is from Properties of Inverse Trigonometric Functions in portion Inverse Trigonometric Functions of Mathematics – Class 12 |
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Answer» RIGHT ANSWER is (a) \(\frac{\sqrt{1-X^2}}{x}\) Easy explanation: LET sin^-1x=y. From ∆ABC, we get y=sin^-1x=cot^-1(\(\frac{\sqrt{1-x^2}}{x}\)) ∴cot(sin^-1x)=\(cot(cot^{-1}(\frac{\sqrt{1-x^2}}{x}))=\frac{\sqrt{1-x^2}}{x}\). |
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