How would you verify the following trig identity $\frac{\sin(x)}{\sin(x)+\cos(x)}=\frac{\sec(x)}{\sec(x)+\cos(x)}$ How to verify the following trig identity.
$$\frac{\sin(x)}{\sin(x)+\cos(x)}=\frac{\sec(x)}{\sec(x)+\cos(x)}$$
I started with the right side and multiplied the numerator and denominator by $\sec(x)-\cos(x)$
then I got
$$\frac{\sec^2(x)-1}{\sec^2(x)-\cos^2(x)}=\frac{\tan^2(x)}{\sec^2(x)-\cos^2(x)}$$
but now I am stuck.
 A: If you could cross multiply this, you will get:
$sin(x)(sec(x)+cos(x))=sec(x)(sin(x)+cos(x))$
$\Rightarrow sin(x)(\frac{1}{cos(x)}+cos(x))=\frac{1}{cos(x)}(sin(x)+cos(x))$
$\Rightarrow \frac{sin(x)}{cos(x)}+sin(x)cos(x)=\frac{sin(x)}{cos(x)}+\frac{cos(x)}{cos(x)}$
$\Rightarrow sin(x)cos(x)=1$
$\Rightarrow 2sin(x)cos(x)=sin(2x)=2$
There's no such $x$ such that it holds. So it is not a valid identity.
A: There may be a typo, either in your question or in your book. Is it possible it was meant to be
$$\frac{\sin(x)}{\sin(x)+\cos(x)}=\frac{\sec(x)}{\sec(x)+\csc(x)}$$
instead (note "$\csc$" instead of "$\cos$")?
If so, you can get this by just dividing each term on the left by $\sin x \cos x$.
A: It's easy to see that the right hand side is $(1+\cos^2x)^{-1}$; so the equality holds if and only if
$$
\sin x(1+\cos^2x)=\sin x+\cos x
$$
that is
$$
\sin x\cos^2x-\cos x=0
$$
Since $\cos x\ne0$ as you need to use the secant, it only remains $\sin x\cos x=1$, which has no solution.
So the equality holds for no value of $x$.
On the other hand
$$
\frac{\sec x}{\sec x+\csc x}=
\frac{\dfrac{1}{\cos x}}{\dfrac{1}{\cos x}+\dfrac{1}{\sin x}}=
\frac{\sin x}{\sin x+\cos x}
$$
So the correct identity is
$$
\frac{\sin x}{\sin x+\cos x}=\frac{\sec x}{\sec x+\csc x}
$$
A: If both sides are  defined (i.e. $\;x\neq \dfrac\pi2, \dfrac{3\pi}4\mod\pi$)
\begin{align*}\frac{\sin x}{\sin x+\cos x}&=\frac{\sec x}{\sec x+\cos x}\iff \sin x(\sec x+\cos x)=(\sin x+\cos x)\sec x\\
&\iff \sin x \cos x=\sec x\cos x\iff\sin x =\sec x\iff\frac12\sin2x=1,\\
\end{align*}
which  has no solution.
A: This is not an identity. Because,when we putting the value of both sides of the identity we get two different value. Left hand value is 0.Right hand value is (1÷2).
