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AP EAPCET 18 May 2024 Shift 1 Solved Paper

Section: Mathematics

Question : 25 of 160

Marks: +1, -0

\cosh^{-1} 2 =

Solution:

To solve this problem, let's use the relationship for the inverse hyperbolic cosine function:

\cosh^{-1} x = \log (x + \sqrt{x^2 - 1})

Applying this to

\cosh^{-1} 2

, we get:

\cosh^{-1} 2 = \log (2 + \sqrt{2^2 - 1})

Simplifying under the square root:

\sqrt{2^2-1} = \sqrt{4-1} = \sqrt{3}

So, the expression becomes:

\cosh^{-1} 2 = \log (2 + \sqrt{3})

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