If f^{prime}(x)=tan ^{-1}(sec x+tan x),-frac{ | vedclass

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(A) Given $f^{\prime}(x) = 	an ^{-1}(\sec x + 	an x)$.Simplifying the expression inside the inverse tangent:$f^{\prime}(x) = 	an ^{-1}\left(\frac{1

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$\frac{\pi+1}{4}$

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(A) Given $f^{\prime}(x) = 	an ^{-1}(\sec x + 	an x)$.Simplifying the expression inside the inverse tangent:$f^{\prime}(x) = 	an ^{-1}\left(\frac{1+\sin x}{\cos x}ight)$Using half-angle identities $\sin x = 2\sin\frac{x}{2}\cos\frac{x}{2}$,$\cos x = \cos^2\frac{x}{2} - \sin^2\frac{x}{2}$,and $1 = \cos^2\frac{x}{2} + \sin^2\frac{x}{2}$:$f^{\prime}(x) = 	an ^{-1}\left[\frac{(\cos \frac{x}{2} + \sin \frac{x}{2})^2}{(\cos \frac{x}{2} - \sin \frac{x}{2})(\cos \frac{x}{2} + \sin \frac{x}{2})}ight]$$f^{\prime}(x) = 	an ^{-1}\left(\frac{\cos \frac{x}{2} + \sin \frac{x}{2}}{\cos \frac{x}{2} - \sin \frac{x}{2}}ight)$Dividing numerator and denominator by $\cos \frac{x}{2}$:$f^{\prime}(x) = 	an ^{-1}\left(\frac{1 + 	an \frac{x}{2}}{1 - 	an \frac{x}{2}}ight) = 	an ^{-1}\left[	an \left(\frac{\pi}{4} + \frac{x}{2}ight)ight]$Thus,$f^{\prime}(x) = \frac{\pi}{4} + \frac{x}{2}$.Integrating with respect to $x$:$f(x) = \int \left(\frac{\pi}{4} + \frac{x}{2}ight) dx = \frac{\pi x}{4} + \frac{x^2}{4} + C$.Given $f(0) = 0$,we get $C = 0$.So,$f(x) = \frac{\pi x + x^2}{4}$.Therefore,$f(1) = \frac{\pi(1) + (1)^2}{4} = \frac{\pi+1}{4}$.

If $f^{\prime}(x)=\tan ^{-1}(\sec x+\tan x),-\frac{\pi}{2} < x < \frac{\pi}{2}$ and $f(0)=0$,then $f(1)$ is

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