If ƒ'(x) = tan–1(secx + tanx), $ - {\pi \over 2} < x < {\pi \over 2}$, and ƒ(0) = 0, then ƒ(1) is equal to :

Question

If ƒ'(x) = tan^–1(secx + tanx), $ - {\pi \over 2} < x < {\pi \over 2}$,
and ƒ(0) = 0, then ƒ(1) is equal to :

${1 \over 4}$

${{\pi - 1} \over 4}$

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

${{\pi + 2} \over 4}$

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