We know that.
#color(violet)((1)cos(pi/2-theta)=sintheta and sin(pi/2-theta)=costheta#
#color(blue)((2) (1+cosalpha)=2cos^2(alpha/2) and sinalpha=2sin(alpha/2)cos(alpha/2)#
#color(brown)((3)cot(pi/2-alpha)=tanalpha#
#color(red)((4)tan^-1(tan alpha)=alpha ,where, alpha in(-pi/2,pi/2)#
Here,
#y=3tan^-1(x+sqrt(1+x^2))#
Substitute, #x=tantheta=>theta=tan^-1x,where,thetain (-
pi/2,pi/2)#
#:.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)#
