Question

Solve the equation

tan^-1(under rootx^2+x)+sin^-1(under rootx^2+x+1)=pi/2

Where tan^-1 inverse of tan

Answer 1

two angles add up to 90 degrees
so they are angles in the same right triangle
the tangent of one of them = opposite /adjacent = a/b
the sin of the other = b/c = b/sqrt(a^2+b^2)
a/b = sqrt (x^2 + x)
b/sqrt(a^2+b^2) = sqrt(x^2 + x + 1)

a^2 = b^2 (x^2+x)= b^2 x^2 + b^2 x

b^2 = (a^2+b^2)(x^2+x+1)

b^2 = b^2 x^2 + b^2 x +b^2)(x^2 + x + 1)
1 = (x^2 + x + 1)(x^2+x+1)
looks like
x^2 + x + 1 = 1
x^2 + x = 0
x(x+1) = 0
x = 0 or x = -1

Answer 2

Thank you

Answer 3

You are welcome. Check arithmetic. I did it fast.