Prove that `tan^(-1)[(sqrt(1+x^2)+sqrt(1-x^2))/(sqrt(1+x^2)-sqrt(1-x^2))]=pi/4+1/2cos^(-1)x^2`

Answer

N/a

Explanation

We have,
`tan^(-1)((sqrt(1+x^(2))+sqrt(1+x^(2)))/(sqrt(1+x^(2))-sqrt(1+x^(2))))=pi/4+1/2cos^(-1)x^(2)`
`:.LHS=tan^(-1)((sqrt(1+x^(2))+sqrt(1-x^(2)))/(sqrt(1+x^(2))-sqrt(1-x^(2))))".........."(i)`
`["let" x^(2)=cos2theta=(cos^(2)theta-sin^(2)theta)=1-2sin^(2)theta=2cos^(2)theta-1]`
`rArr cos^(-1)x^(2)=2theta rArr theta = 1/2 cos^(-1)x^(2)`
`:. sqrt(1+x^(2)) = sqrt(1+cos2theta)`
`= sqrt(1+2cos^(2)theta-1) = sqrt(2) cos theta`
and `sqrt(1-x^(2))=sqrt(1-cos2theta)`
`= sqrt(1+2cos^(2)theta) = sqrt(2) sin theta`
`:. LHS =tan^(-1)((sqrt(2)costheta + sqrt(2)sintheta)/(sqrt(2)costheta-sqrt(2)sintheta))`
`=tan^(-1)((costheta+sintheta)/(costheta-sintheta))`
`= tan^(-1)((1+sintheta)/(1-tantheta))=tan^(-1)((tan'(pi)/(4)+tantheta)/(1-tan'(pi)/(4).tantheta))`
`tan^(-1)[tan(pi/4+theta)] , [:' tan(x+y)=(tanx+tany)/(1-tanx.tany)]`
`= pi/4+theta=pi/4+1/2costheta^(-1)x^(2)`
`=RHS`