\frac { d } { d x } \tan ^ { - 1 } \left( \frac { \sqrt { 1 + \sin x }

1.	\frac { d } { d x } \tan ^ { - 1 } \left( \frac { \sqrt { 1 + \sin x } + \sqrt { 1 + \sin x } } { \sqrt { 1 + \sin x } - \sqrt { 1 + \sin x } } \right) =
Answer» tan-1√1+sinx/1-sinsx^ = tan-1√(cos^2x/2 + sin^2x/2 + 2sinx/2cosx/2)/(cos^2x/2 + sin^2x/2 – 2sinx/2cosx/2) =tan-1√(cosx/2 + sinx/2)^2/(cosx/2–sinx/2)2 =tan-1(cosx/2 + sinx/2)/(cosx/2–sinx/2) =tan-1(1 + tanx/2)/(1–tanx/2) (by dividing the numerator & denominator by cos x/2) =tan-1(tan(pi/4) + x/2) =(pi/4) + x/2 Henced/dx tan-1√1+sinx/1-sinx = d/dx [(pi/4) + x/2] = 1/2 Ans: 1/2

\frac { d } { d x } \tan ^ { - 1 } \left( \frac { \sqrt { 1 + \sin x } + \sqrt { 1 + \sin x } } { \sqrt { 1 + \sin x } - \sqrt { 1 + \sin x } } \right) =

Answer»

tan-1√1+sinx/1-sinsx^

= tan-1√(cos^2x/2 + sin^2x/2 + 2*sinx/2*cosx/2)/(cos^2x/2 + sin^2x/2 – 2*sinx/2*cosx/2)

=tan-1√(cosx/2 + sinx/2)^2/(cosx/2–sinx/2)2

=tan-1(cosx/2 + sinx/2)/(cosx/2–sinx/2)

=tan-1(1 + tanx/2)/(1–tanx/2) (by dividing the numerator & denominator by cos x/2)

=tan-1(tan(pi/4) + x/2)

=(pi/4) + x/2

Henced/dx tan-1√1+sinx/1-sinx = d/dx [(pi/4) + x/2] = 1/2

Ans: 1/2