If f'(x)=\tan ^{2}(x)+\cot ^{2}(x) and f(\frac{\\pi}{4})=0, then f(x)=

Correct Answer is : tan(x)−cot(x)−2x+π2

\tan (x)-\cot (x)-x+\frac{\\pi}{2}

\tan (x)-\cot (x)-2x+\frac{\\pi}{2}

\tan (x)+\cot (x)-2x+\frac{\\pi}{2}

\sec(x)-\cosec(x)-2x+\frac{\\pi}{2}

AP EAPCET 23-Aug-2021 Shift 2 Paper

Section: Mathematics

Question : 76 of 160

Marks: +1, -0

If f′(x)=tan2(x)+cot2(x) and f(

)=0, then f(x)=

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