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)#
