How to write as an algebraic expression

Question

We have cot(cos^-1 √x²-1/x) For x > 1

I drew a triangle and used the Pythagorean theorem. However, I did x²-(√x²-1)² = b² and I got x²-x²-1 = b² where am I going wrong? Thank you!

Answer 1

x²-(√x²-1)² = b²

^{x^2 -( x^2-1)^2 =b^2}

x^2 -x^2 +1= b^2

1=b^2

is the algebra that went wrong...

Answer 2

cos^-1 √x²-1/x) For x > 1

IS A TRIANGLE (OR SIMILAR TRIANGLE)

with x= √x²-1/x r=1 and y= √(1-(x²-1)/x²)=1/x where r is hypontenuse and x is adjacent side and y opposite side of the angle cos^-1 √((x²-1)/x)

cot (cos^-1 √(x²-1)/x)=[√(x²-1)/x]/[1/x]=cot (cos^-1 √(x²-1)/x)

question is as function of x, no trigonometry, what is f)x)=cot (cos^-1 √(x²-1)/x)

f(x)=√(x²-1) x>1