2305Given expression :
√1+x/a - √1-x/a√1+x/a + √1-x/a
Putting cos(2θ) = x/a and making the substitutions,
1+cos(2θ) = 2cos2(θ) and 1-cos(2θ) = 2sin2(θ)
We get the expression to be:
cos(θ)-sin(θ)cos(θ)+sin(θ) = tan(π/4 - θ)
dθ/dx = d(1/2 * cos-1x/a)dx = 12√a2-x2
Therefore dtan(π/4 - θ)/dx = -sec2(π/4 - 1/2*cos-1x/a)2√a2-x2
