We know that,
color(blue)((1)1-costheta=2sin^2(theta/2) and sintheta=2sin(theta/2)cos(theta/2)(1)1−cosθ=2sin2(θ2)andsinθ=2sin(θ2)cos(θ2)
color(violet)((2)arc tan (-alpha)=-arc tanalpha(2)arctan(−α)=−arctanα
color(green)((3)arc tan(tanphi)=phi ,where,phi in(-pi/2,pi/2)(3)arctan(tanϕ)=ϕ,where,ϕ∈(−π2,π2)
Let
y=arc tan(x-sqrt(1+x^2))y=arctan(x−√1+x2)
We take, color(brown)(x=cottheta=>theta=arc cotx,where,theta in(0,pi)=>theta/2in(0,pi/2)x=cotθ⇒θ=arccotx,where,θ∈(0,π)⇒θ2∈(0,π2)
So,
y=arc tan(cottheta-sqrt(1+cot^2theta))to[use : color(red)(1+cot^2theta=csc^2theta]y=arctan(cotθ−√1+cot2θ)→[use:1+cot2θ=csc2θ
=>y=arc tan(costheta/sintheta-csctheta)⇒y=arctan(cosθsinθ−cscθ)
=>y=arc tan (costheta/sintheta-1/sintheta)⇒y=arctan(cosθsinθ−1sinθ)
=>y=arc tan((costheta-1)/sintheta)...tocolor(violet)(Apply(2)
=>y=arc tan{-((1-costheta)/sintheta)}
=>y=-arc tan((1-costheta)/sintheta)...tocolor(blue)(Apply(1)
=>y=-arc tan((2sin^2(theta/2))/(2sin(theta/2)cos(theta/2)))
=>y=-arc tan((sin(theta/2))/cos(theta/2))
=>y=-arc tan(tan(theta/2)).tocolor(green)(Apply(3) ,as ,theta/2in(0,pi/2)
=>y=-theta/2
Subst. back , color(brown)(theta=arc cotx
y=-1/2*arc cotx
=>(dy)/(dx)=-1/2*(-1/(1+x^2))
=>(dy)/(dx)=1/(2(1+x^2)
