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differentiate tan^-1(√1-x^2)/(√1+x^2) w.r.t cos^-1 x
Let tan^-1{√1+x^2 -√1-x^2 /√1+x^2 + √1-x^2}=uLet x^2=cos2t and thentan u = √1+cos2t - √1-cos2t / √1+cost +√1-cos2ttan u = √1+2cos^2t-1 - √1-1+2sin^2t / √1+2cos^2t-1 + √1-1+2sin^2ttan u = cost + sint / cost - sintDivided by cost tan u = 1+tant /1-tantu= tan^-1{tan(t+π/4)}u=(t+π/4)t=1/2 cos^-1 x^2u=π/4 + 1/2 cos^-1 x^2du/dx=-x/√1-x^4Cos^-1 x^2 =vdv/dx=-2x/√1-x^4du/dv=du/dx * dx/dvdu/dv=1/2u=π/4 + 1/2 cos^-1 x^2du/dx=1/2(-1/√1-x^
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