If f^{prime}(x)=tan ^2(x)+cot ^2(x) and fleft | vedclass

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(B) Given $f^{\prime}(x)=	an ^2 x+\cot ^2 x$. Integrating both sides with respect to $x$: $f(x)=\int(	an ^2 x+\cot ^2 x) dx$. Using the identity $

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) Given $f^{\prime}(x)=	an ^2 x+\cot ^2 x$. Integrating both sides with respect to $x$: $f(x)=\int(	an ^2 x+\cot ^2 x) dx$. Using the identity $	an^2 x = \sec^2 x - 1$ and $\cot^2 x = \operatorname{cosec}^2 x - 1$: $f(x)=\int(\sec^2 x - 1 + \operatorname{cosec}^2 x - 1) dx$. $f(x)=\int(\sec^2 x + \operatorname{cosec}^2 x - 2) dx$. $f(x)=	an x - \cot x - 2x + C$. Given $f\left(\frac{\pi}{4}ight)=0$: $f\left(\frac{\pi}{4}ight)=	an\left(\frac{\pi}{4}ight) - \cot\left(\frac{\pi}{4}ight) - 2\left(\frac{\pi}{4}ight) + C = 0$. $1 - 1 - \frac{\pi}{2} + C = 0$. $C = \frac{\pi}{2}$. Therefore,$f(x)=	an x - \cot x - 2x + \frac{\pi}{2}$.

If $f^{\prime}(x)=\tan ^2(x)+\cot ^2(x)$ and $f\left(\frac{\pi}{4}\right)=0$,then $f(x)$ is equal to:

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