I hope the **Question** is :

If, #sin^-1x+tan^-1x=pi/2#, prove that, #2x^2+1=5#.

Let #tan^-1x=theta :. tan theta=x......(star1)#.

Given that, #sin^-1x+tan^-1x=pi/2 :. sin^-1x+theta=pi/2#.

#:. sin^-1x=pi/2-theta#.

#:. sin(pi/2-theta)=x#.

#:. costheta=x, or, sectheta=1/x......(star2)#.

#"But, "tan^2theta=sec^2theta-1#.

#:. x^2=1/x^2-1, or, x^4+x^2-1=0#.

This is a **quadr. eqn.** in #x^2#.

Solving it with the help of the **quadr. formula,** we get,

#x^2={-1+-sqrt(1+4)}/2=(-1+-sqrt5)/2#.

Since, #x^2 >0, x^2!=(-1-sqrt5)/2#.

#:. x^2=(-1+sqrt5)/2#, or what is the same as to say that,

# 2x^2+1=sqrt5#.