contestada

How do I factor the algebraic expression, csc^2(x) - 1, in terms of a single trigonometric function?

Respuesta :

[tex]\csc^2(x)-1=\left(\dfrac{1}{\sin(x)}\right)^2-1=\dfrac{1}{\sin^2(x)}-1\\\\=\dfrac{1}{\sin^2(x)}-\dfrac{\sin^2(x)}{\sin^2(x)}=\dfrac{1-\sin^2(x)}{\sin^2(x)}=\dfrac{\cos^2(x)}{\sin^2(x)}=\left(\dfrac{\cos(x)}{\sin(x)}\right)^2\\\\=\cot^2(x)\\\\\boxed{\csc^2(x)-1=\cot^2(x)}\\\\Used:\\\\\csc(x)=\dfrac{1}{\sin(x)}\\\\\sin^2(x)+\cos^2(x)=1\to\cos^2(x)=1-\sin^2(x)\\\\\dfrac{\cos(x)}{\sin(x)}=\cot(x)[/tex]