At x=\frac{\pi^2}{4}, \frac{d}{d x} ({\Tan}^{-1}(\cos \sqrt{ {\x})+{\Sec}^{-1} ( {\e}^{ {\x}}) ) =

Correct Answer is : 1√eπ22−1−1π

{1}/\sqrt{ {\e}^{\frac{\pi^2}{2}}}-1}-\frac{1}{\pi}

\frac{\pi}{4}+ {1}/\sqrt{ {\e}^{\pi^2}+ {\e}^{\pi^2 \/ 2}}}

{1}/\sqrt{ {\e}^{\pi^2}+ {\e}^{\pi^2 \/ 2}}}+\frac{2}{\pi} \cot ( {\sqrt{\pi}}/{2})

{1}/\sqrt{ {\e}^\pi}}+\frac{1}{\pi}

AP EAPCET 19th May 2023 Shift 1 Solved Paper

Section: Mathematics

Question : 62 of 160

Marks: +1, -0

At x=‌

π2

,‌

(Tan−1(cos‌√x)+Sec−1(ex))=

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