TS EAMCET · Maths · Inverse Trigonometric Functions
\(x=\log \left(\frac{1}{y}+\sqrt{1+\frac{1}{y^2}}\right) \Rightarrow y\) is equal to
- A \(\tanh x\)
- B \(\operatorname{coth} x\)
- C \(\operatorname{sech} x\)
- D \(\operatorname{cosech} x\)
Answer & Solution
Correct Answer
(D) \(\operatorname{cosech} x\)
Step-by-step Solution
Detailed explanation
Given, \(x=\log \left(\frac{1}{y}+\sqrt{1+\frac{1}{y^2}}\right)\) \(\begin{array}{ll}\therefore & x=\operatorname{cosech}^{-1} y \\ \Rightarrow & y=\operatorname{cosech} x\end{array}\)
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