Is $\sec(x+y) =\frac{\cos(x+y)} {\cos^2 x} $ an identity?

Question

Is $$\sec(x+y) =\frac{\cos(x+y)} {\cos^2 x} $$ a trigonometric identity? If yes, how could it be proved? I tried very hard but it seems to be difficult

Answer 1

$\sec(x+y) =\frac{\cos(x+y)} {\cos^2 x} \iff $

$\frac 1{\cos (x+y)} = \frac{\cos(x+y)} {\cos^2 x} \iff$

$\cos(x+y) = \frac {\cos^2 x}{\cos(x+y)} \iff $

$\cos^2(x+y) = \cos^2 x$

Which is obviously not the case (consider $x= 0$ and $\cos y \ne \pm 1$ or ...pretty much any $y \ne 0,-2x, \pi - 2x, \pi$).

So, no, it is not an identity.

Answer 2

Hint:

$$\cos^2(x+y)=\cos^2x\iff1-\sin^2(x+y)=1-\sin^2x$$

Using Prove $ \sin(A+B)\sin(A-B)=\sin^2A-\sin^2B $,

$$0=\sin^2(x+y)-\sin^2x=\sin(2x+y)\sin y$$

Answer 3

Counterexample

Let $x=0$ and $y=\pi/3$. Then $$\text{LHS}=\sec\left(0+\frac\pi3\right)=\frac1{\cos(\pi/3)}=2$$ whereas $$\text{RHS}=\frac{\cos(0+\pi/3)}{\cos^20}=\frac12$$ so $\text{LHS}\neq\text{RHS}$.