Simplify :

`1/sqrt2 + pi/4 ( 1/sqrt2 ) + 1/sqrt2 - pi/4 ( 1/sqrt2 )`

Simplify (1/2+1/sqrt2-sqrt3)/(sqrt3-1/sqrt2-1/sqrt3)

Show that tan pi/16 = sqrt(4 +2 sqrt2) - (sqrt2+1) .

The value of { 1/(sqrt6 - sqrt5) - 1/(sqrt5 - sqrt4) + 1/(sqrt4 - sqrt3) - 1/(sqrt3 - sqrt2) + 1/(sqrt2 - 1)} is :

The function y=f(x) is the solution of the differential equation [dy]/[dx]+[xy]/[x^2-1]=[x^4+2x]/sqrt[1-x^2] in (-1, 1), satisfying f(0)=0. Then int_[-sqrt3/2]^[sqrt3/2] f(x)dx is (A) pi/3 - sqrt3/2 (B) pi/3 - sqrt3/4 (C) pi/6 - sqrt3/4 (D) pi/6 - sqrt3/2

Simplify : (1)/(sqrt(3)+sqrt(2))-(1)/(sqrt(3)-sqrt(2))+(2)/(sqrt(2)+1)

If (dy)/(dx) +y sec x=tan x," then (sqrt(2) +1) y((pi)/(4)) - y(0)=, (A) sqrt(2) - (pi)/(4) (B) sqrt(2) + (pi)/(4) (C) sqrt(2) - (pi)/(2) (D) sqrt(2) + (pi)/(2)

Simplify 1/sqrt2+1/(sqrt2+sqrt4)+1/(sqrt4+sqrt6)+1/(sqrt6+sqrt8)

Simplify: (1)/(1+sqrt(2))+(1)/(sqrt(2)+sqrt(3))+(1)/(sqrt(3)+sqrt(5))

Simplify sqrt((sqrt(2)+1)/(sqrt(2)-1))