If x ∊ ≤ft(-π/2, π/2) , then log sec x =

Correct Answer is : 2 coth⁡−1(csc⁡2x2−1)^{-1}≤ft(² x/2 - 1)coth−1(csc22x−1)

Correct Answer is : 2 coth⁡−1(csc⁡2x2−1)^{-1}≤ft(² x/2 - 1)coth−1(csc22x−1)

Test Index

AP EAMCET Engineering 21 Apr 2019 Shift 1 Paper

Section: Mathematics

Question : 25 of 160

Marks: +1, -0

If x ∊

\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)

, then log sec x =

Solution:

\;\text{ For }\;, x \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)

\;\text{ log sec }\; x=y \;\text{ (let) }\;

Then

\sec x \;\; = e^{x}

\cos x \;\; = e^{y}

\;\frac{\sinh y}{\cosh y} \;\; = \;\frac{e^{y}-e^{-y}}{e^{y}+e^{-y}}

\;\frac{\cosh y-\sinh y}{\cosh y+\sinh y} \;\; = \;\frac{e^{-y}}{e^{y}}

\;\; = \cos^{2} x

Solve further

\left(\;\frac{\cosh\frac{y}{2}-\sinh\frac{y}{2}}{\cosh\frac{y}{2}+\sinh\frac{y}{2}}\right)^{2} = \cos^{2} x

\;\frac{\cosh\frac{y}{2}-\sinh\frac{y}{2}}{\cosh\frac{y}{2}+\sinh\frac{y}{2}} = \cos x

\;\frac{2\cosh\frac{y}{2}}{2\sinh\frac{y}{2}} \;\; = \;\frac{1+\cos x}{1-\cos x}

\;\; = \cos^{2} \frac{x}{2}

Simplify further,

\coth \frac{y}{1} \;\; = \cot^{2} \frac{x}{2}

\;\; = \csc \frac{x}{2} - 1

\;\frac{y}{2} \;\; = \coth^{-1}\left(\csc^{2} \frac{x}{2} - 1\right)

y \;\; = 2 \coth^{-1}\left(\csc^{2} \frac{x}{2} - 1\right)

Go to Question:

Prev Question

Next Question