Let #u = Sin^(-1)x#.

Let #v = Tan^(-1)x#.

Since #sinu = x#, we may draw a right triangle having #x# as the length of the side opposite angle u, and #1# as the length of the hypotenuse. This we do because #sin = (opp)/(hyp)#. By the Pythagorean Theorem, the length of the remaining leg of that triangle is #sqrt(1 - x^2)#.

Since #tanv = x#, we may draw a right triangle having #x# as the length of the side opposite angle v, and #1# as the length of the side adjacent to angle v. This we do because #tan = (opp)/(adj)#. By the Pythagorean Theorem, the length of the remaining leg of that triangle is #sqrt(1 + x^2)#.

Now examine #Sin^(-1)x + Tan^(-1)x = pi/2#.

We take the cosine of both sides, since #cos(pi/2) = 0#.

#cos(Sin^(-1)x + Tan^(-1)x) = cos(pi/2)#

#cos(Sin^(-1)x + Tan^(-1)x) = 0#

Using u and v, we have:

#cos(u + v) = 0#

Apply the cosine addition formula:

#cosucosv-sinusinv = 0#

From the first triangle that we drew, above, we see that

#cosu = sqrt(1 - x^2)# and #sinu = x#.

From the second triangle we see that

#cosv = 1/sqrt(1 + x^2)# and #sinv = x/sqrt(1 + x^2)#.

The addition formula gives us:

#cosucosv-sinusinv = 0#

#(sqrt(1 - x^2))(1/sqrt(1 + x^2)) - (x)(x/sqrt(1 + x^2)) = 0#

#sqrt(1 - x^2)/sqrt(1 + x^2) - x^2/sqrt(1 + x^2) = 0#

Multiply by the denominator, which is nonzero:

#sqrt(1 - x^2) - x^2 = 0#

Now for some algebra:

#sqrt(1 - x^2) = x^2#

#1 - x^2 = x^4#

#0= x^4 + x^2 - 1#

By the Quadratic Formula,

#x^2 = (-1 pm sqrt(5))/2#

Since #x^2# must be positive, replace #pm# with +

#x^2 = (-1 + sqrt(5))/2#

So that x has exactly the values described in the earlier solution.

Now, the values of both #arcsinx# and #arctanx# are negative when #x < 0#. Therefore, the negative root is not a solution to the equation. The only solution is:

#x = sqrt((-1 + sqrt(5))/2)#