Question

Please help!

Verify each trigonometric equation by substituting identities to match the right hand side of the equation to the left hand side of the equation.

- tan2x + sec2x = 1

Answer 1

We're gonna need some definitions here\[\tan(x) = \frac{ \sin(x) }{ \cos(x) }, \; \sec(x) = \frac{ 1 }{ \cos(x) }\]\[ -\tan(x)^2 + \sec(x)^2 = -\frac{\sin(x)^2}{\cos(x)^2} + \frac{1}{\cos(x)^2} = 1\]\[\frac{1-\sin(x)^2}{\cos(x)^2} = 1 \Leftrightarrow 1-\sin(x)^2 = \cos(x)^2 \Leftrightarrow 1 = \cos(x)^2 + \sin(x)^2\]Which is true for all real x.

Answer 2

Can you explain it more please? I really dont understand this

Answer 3

Where's the problem? Should I explain why cos(x)^2 + sin(x)^2 = 1?

Answer 4

Im not sure the it says to verify the equation by substituting identities to match the right hand side of the equation to the left side of the equation

Answer 5

yes, that is exactly what I've done. I've used the identity\[\tan(x) = \frac{ \sin(x) }{ \cos(x) }, \; \] as well as the identity \[\sec(x) = \frac{ 1 }{ \cos(x)} \] along with the property \[\sin(x)^2 + \cos(x)^2 = 1\] I've substituted them into our original statement, did some transforms and got a true statement, which implies that our original statement is also true.

Answer 6

Ohh okay