We know that.
color(violet)((1)cos(pi/2-theta)=sintheta and sin(pi/2-theta)=costheta(1)cos(π2−θ)=sinθandsin(π2−θ)=cosθ
color(blue)((2) (1+cosalpha)=2cos^2(alpha/2) and sinalpha=2sin(alpha/2)cos(alpha/2)(2)(1+cosα)=2cos2(α2)andsinα=2sin(α2)cos(α2)
color(brown)((3)cot(pi/2-alpha)=tanalpha(3)cot(π2−α)=tanα
color(red)((4)tan^-1(tan alpha)=alpha ,where, alpha in(-pi/2,pi/2)(4)tan−1(tanα)=α,where,α∈(−π2,π2)
Here,
y=3tan^-1(x+sqrt(1+x^2))y=3tan−1(x+√1+x2)
Substitute, x=tantheta=>theta=tan^-1x,where,thetain (-
pi/2,pi/2)x=tanθ⇒θ=tan−1x,where,θ∈(−π2,π2)
:.y=3tan^-1(tantheta+sqrt(1+tan^2theta))
=3tan^-1(tantheta+sectheta)
=3tan^-1(sintheta/costheta+1/costheta)
=3tan^-1((1+sintheta)/costheta)..to Apply(1)
=3tan^-1((1+cos(pi/2-theta))/sin(pi/2-theta))...toApply(2)
=3tan^-1((2cos^2(pi/4-theta/2))/(2sin(pi/4-theta/2)cos(pi/4-
theta/2)))
=3tan^-1(cos(pi/4-theta/2)/sin(pi/4-theta/2))
=3tan^-1(cot(pi/4-theta/2))
=3tan^-1(cot{pi/2-pi/4-theta/2})
=3tan^-1(cot(pi/2-(pi/4+theta/2))...toApply(3)
=3tan^-1(tan(pi/4+theta/2))..toApply(4)
Now,
theta in (-pi/2,pi/2)
=>theta/2 in(-pi/4,pi/4)...to [Dividing by 2]
=>pi/4+theta/2 in (-pi/4+pi/4,pi/4+pi/4)...to[Adding pi/4]
=>pi/4+theta/2 in(0,pi/2)sub(-pi/2, pi/2)
So,
y=3(pi/4+theta/2)
y=(3pi)/4+3/2*theta,where, theta=tan^-1x
=>y=(3pi)/4+3/2tan^-1x
=>(dy)/(dx)=0+3/2(1/(1+x^2))
i.e. (dy)/(dx)=3/(2(1+x^2)
