1.

What is the solution of cot⁡(sin^-1⁡x)?(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

Answer» RIGHT ANSWER is (a) \(\frac{\sqrt{1-X^2}}{x}\)

Easy explanation: LET sin^-1⁡x=y. From ∆ABC, we get

y=sin^-1⁡x=cot^-1⁡(\(\frac{\sqrt{1-x^2}}{x}\))

∴cot⁡(sin^-1⁡x)=\(cot⁡(cot^{-1}(\frac{\sqrt{1-x^2}}{x}))=\frac{\sqrt{1-x^2}}{x}\).


Discussion

No Comment Found

Related InterviewSolutions