Answer:
[tex]\boxed{sin^2x+cos^2x=1}[/tex]
Step-by-step explanation:
We want to verify that;
[tex]\sin(x) \cos(x) \tan(x)=1-\cos^2(x)[/tex]
We take only the expression on the left hand side and work to get the expression on the right hand side.
[tex]\sin(x) \cos(x) \tan(x)[/tex]
[tex]=\sin(x) \cos(x) \frac{\sin(x)}{\cos(x)}[/tex]
We cancel the common factors to get;
[tex]=\sin(x) \times \sin(x)[/tex]
[tex]=\sin^2(x)[/tex]
We now use the Pythagorean identity;
[tex]sin^2x+cos^2x=1[/tex]
We make [tex]sin^2x[/tex] the subject to obtain;
[tex]sin^2x=1-cos^2x[/tex]
as required.
The basic trigonometric identity we used is [tex]sin^2x+cos^2x=1[/tex]
