Answer
Some of the identities that are verified using sum-to product formulas are:
Verified identities are:
$\begin{align}
& \tan x=\frac{\cos 3x-\cos 5x}{\sin 3x+\sin 5x} \\
& -\tan x=\frac{\cos 3x-\cos x}{\sin 3x+\sin x} \\
& -\cot 2x=\frac{\sin 3x-\sin x}{\cos 3x-\cos x} \\
& \tan 3x=\frac{\sin 2x+\sin 4x}{\cos 2x+\cos 4x}
\end{align}$
Other identities are:
$\begin{align}
& \frac{\sin x-\sin y}{\sin x+\sin y}=\tan \frac{x-y}{2}\cot \frac{x+y}{2} \\
& \frac{\sin x+\sin y}{\sin x-\sin y}=\tan \frac{x+y}{2}\cot \frac{x-y}{2} \\
& \frac{\sin x-\sin y}{\cos x-\cos y}=-\cot \frac{x+y}{2}
\end{align}$
And,
$\begin{align}
& \frac{\sin x+\sin 3x}{\cos x+\cos 3x}=\tan 2x \\
& \frac{\cos 4x-\cos 2x}{\sin 2x-\sin 4x}=\tan 3x
\end{align}$
Sum-to product formulas are given below:
$\begin{align}
& \sin \alpha +\sin \beta =2\sin \frac{\alpha +\beta }{2}\cos \frac{\alpha -\beta }{2} \\
& \sin \alpha -\sin \beta =2\sin \frac{\alpha -\beta }{2}\cos \frac{\alpha +\beta }{2} \\
& \cos \alpha +\cos \beta =2\cos \frac{\alpha +\beta }{2}\cos \frac{\alpha -\beta }{2} \\
& \cos \alpha -\cos \beta =-2\sin \frac{\alpha +\beta }{2}\sin \frac{\alpha -\beta }{2}
\end{align}$
Work Step by Step
Let us consider the identity:
$\frac{\cos 3x-\cos 5x}{\sin 3x+\sin 5x}=\tan x$
By using the sum-to product formula $\cos \alpha -\cos \beta =-2\sin \frac{\alpha +\beta }{2}\sin \frac{\alpha -\beta }{2}$ to numerator and sum-to product formula $\sin \alpha +\sin \beta =2\sin \frac{\alpha +\beta }{2}\cos \frac{\alpha -\beta }{2}$ to denominator of left hand side.
Then,
$\begin{align}
& \frac{\cos 3x-\cos 5x}{\sin 3x+\sin 5x}=\frac{-2\sin \frac{3x+5x}{2}\sin \frac{3x-5x}{2}}{2\sin \frac{3x+5x}{2}\cos \frac{3x-5x}{2}} \\
& =\frac{-2\sin \frac{8x}{2}\sin \frac{-2x}{2}}{2\sin \frac{8x}{2}\cos \frac{-2x}{2}} \\
& =\frac{-2\sin 4x\sin \left( -x \right)}{2\sin 4x\cos \left( -x \right)} \\
& =-\tan \left( -x \right)
\end{align}$
Thus, the result will be $\tan x$. Hence, verified.
Also, consider the identity:
$\frac{\cos 3x-\cos x}{\sin 3x+\sin x}=-\tan x$
And use the sum-to product formula $\cos \alpha -\cos \beta =-2\sin \frac{\alpha +\beta }{2}\sin \frac{\alpha -\beta }{2}$ to numerator and sum-to product formula $\sin \alpha +\sin \beta =2\sin \frac{\alpha +\beta }{2}\cos \frac{\alpha -\beta }{2}$ to denominator of left hand side. Then,
$\begin{align}
& \frac{\cos 3x-\cos x}{\sin 3x+\sin x}=\frac{-2\sin \frac{3x+x}{2}\sin \frac{3x-x}{2}}{2\sin \frac{3x+x}{2}\cos \frac{3x-x}{2}} \\
& =\frac{-2\sin \frac{4x}{2}\sin \frac{2x}{2}}{2\sin \frac{4x}{2}\cos \frac{2x}{2}} \\
& =\frac{-2\sin 2x\sin x}{2\sin 2x\cos x} \\
& =-\tan x
\end{align}$
Hence verified.
Also, consider the identity:
$\frac{\sin 3x-\sin x}{\cos 3x-\cos x}=-\cot 2x$
By using the sum-to product formula $\sin \alpha -\sin \beta =2\sin \frac{\alpha -\beta }{2}\cos \frac{\alpha +\beta }{2}$ to numerator and $\cos \alpha -\cos \beta =-2\sin \frac{\alpha +\beta }{2}\sin \frac{\alpha -\beta }{2}$ to denominator of left hand side.
Then,
$\begin{align}
& \frac{\sin 3x-\sin x}{\cos 3x-\cos x}=\frac{2\sin \frac{3x-x}{2}\cos \frac{3x+x}{2}}{-2\sin \frac{3x+x}{2}\sin \frac{3x-x}{2}} \\
& =\frac{2\sin \frac{2x}{2}\cos \frac{4x}{2}}{-2\sin \frac{4x}{2}\sin \frac{2x}{2}} \\
& =-\frac{\cos 2x}{\sin 2x} \\
& =-\cot 2x
\end{align}$
Hence verified.
And also consider the identity:
$\frac{\sin x-\sin y}{\sin x+\sin y}=\tan \frac{x-y}{2}\cot \frac{x+y}{2}$
Also, use the sum-to product formula $\sin \alpha -\sin \beta =2\sin \frac{\alpha -\beta }{2}\cos \frac{\alpha +\beta }{2}$ to numerator and $\sin \alpha +\sin \beta =2\sin \frac{\alpha +\beta }{2}\cos \frac{\alpha -\beta }{2}$ to denominator of left hand side.
Then,
$\begin{align}
& \frac{\sin x-\sin y}{\sin x+\sin y}=\frac{2\sin \frac{x-y}{2}\cos \frac{x+y}{2}}{2\sin \frac{x+y}{2}\cos \frac{x-y}{2}} \\
& =\tan \frac{x-y}{2}\cot \frac{x+y}{2}
\end{align}$
Hence verified.
Also, consider the identity:
$\frac{\sin x+\sin y}{\sin x-\sin y}=\tan \frac{x+y}{2}\cot \frac{x-y}{2}$
And use the sum-to product formula $\sin \alpha -\sin \beta =2\sin \frac{\alpha -\beta }{2}\cos \frac{\alpha +\beta }{2}$ to denominator and $\sin \alpha +\sin \beta =2\sin \frac{\alpha +\beta }{2}\cos \frac{\alpha -\beta }{2}$ to numerator of left hand side.
Then,
$\begin{align}
& \frac{\sin x+\sin y}{\sin x-\sin y}=\frac{2\sin \frac{x+y}{2}\cos \frac{x-y}{2}}{2\sin \frac{x-y}{2}\cos \frac{x+y}{2}} \\
& =\tan \frac{x+y}{2}\cot \frac{x-y}{2}
\end{align}$
Hence verified.
And consider the identity:
$\frac{\sin x+\sin y}{\cos x+\cos y}=\tan \frac{x+y}{2}$
Then use the sum-to product formula $\sin \alpha +\sin \beta =2\sin \frac{\alpha +\beta }{2}\cos \frac{\alpha -\beta }{2}$ to numerator and $\cos \alpha +\cos \beta =2\cos \frac{\alpha +\beta }{2}\cos \frac{\alpha -\beta }{2}$ to denominator of left hand side. Then,
$\begin{align}
& \frac{\sin x+\sin y}{\cos x+\cos y}=\frac{2\sin \frac{x+y}{2}\cos \frac{x-y}{2}}{2\cos \frac{x+y}{2}\cos \frac{x-y}{2}} \\
& =\tan \frac{x+y}{2}
\end{align}$
Hence verified.
Also, consider the identity:
$\frac{\sin x-\sin y}{\cos x-\cos y}=-\cot \frac{x+y}{2}$
And use the sum-to product formula $\sin \alpha -\sin \beta =2\sin \frac{\alpha -\beta }{2}\cos \frac{\alpha +\beta }{2}$ to numerator and $\cos \alpha -\cos \beta =-2\sin \frac{\alpha +\beta }{2}\sin \frac{\alpha -\beta }{2}$ to denominator of left hand side.
Then,
$\begin{align}
& \frac{\sin x-\sin y}{\cos x-\cos y}=\frac{2\sin \frac{x-y}{2}\cos \frac{x+y}{2}}{-2\sin \frac{x+y}{2}\sin \frac{x-y}{2}} \\
& =-\cot \frac{x+y}{2}
\end{align}$
Hence verified.
Also, consider the identity;
$\frac{\sin 2x+\sin 4x}{\cos 2x+\cos 4x}=\tan 3x$
And use the sum-to product formula $\sin \alpha +\sin \beta =2\sin \frac{\alpha +\beta }{2}\cos \frac{\alpha -\beta }{2}$ to numerator and $\cos \alpha +\cos \beta =2\cos \frac{\alpha +\beta }{2}\cos \frac{\alpha -\beta }{2}$ to denominator of left hand side.
Then,
$\begin{align}
& \frac{\sin 2x+\sin 4x}{\cos 2x+\cos 4x}=\frac{2\sin \frac{2x+4x}{2}\cos \frac{2x-4x}{2}}{2\cos \frac{2x+4x}{2}\cos \frac{2x-4x}{2}} \\
& =\frac{2\sin \frac{6x}{2}\cos \frac{-2x}{2}}{2\cos \frac{6x}{2}\cos \frac{-2x}{2}} \\
& =\frac{\sin 3x}{\cos 3x} \\
& =\tan 3x
\end{align}$
Hence verified.
Also, consider the identity:
$\frac{\sin x+\sin 3x}{\cos x+\cos 3x}=\tan 2x$
And use the sum-to product formula $\sin \alpha +\sin \beta =2\sin \frac{\alpha +\beta }{2}\cos \frac{\alpha -\beta }{2}$ to numerator and $\cos \alpha +\cos \beta =2\cos \frac{\alpha +\beta }{2}\cos \frac{\alpha -\beta }{2}$ to denominator of left hand side.
Then,
$\begin{align}
& \frac{\sin x+\sin 3x}{\cos 3x+\cos 3x}=\frac{2\sin \frac{x+3x}{2}\cos \frac{x-3x}{2}}{2\cos \frac{x+3x}{2}\cos \frac{x-3x}{2}} \\
& =\frac{2\sin \frac{4x}{2}\cos \frac{-2x}{2}}{2\cos \frac{4x}{2}\cos \frac{-2x}{2}} \\
& =\frac{\sin 2x}{\cos 2x} \\
& =\tan 2x
\end{align}$
Hence verified.
Also, consider the identity:
$\frac{\cos 4x-\cos 2x}{\sin 2x-\sin 4x}=\tan 3x$
And use the sum-to product formula $\sin \alpha -\sin \beta =2\sin \frac{\alpha -\beta }{2}\cos \frac{\alpha +\beta }{2}$ to denominator and $\cos \alpha -\cos \beta =-2\sin \frac{\alpha +\beta }{2}\sin \frac{\alpha -\beta }{2}$ to numerator of left hand side.
Then,
$\begin{align}
& \frac{\cos 4x-\cos 2x}{\sin 2x-\sin 4x}=\frac{-2\sin \frac{4x+2x}{2}\sin \frac{4x-2x}{2}}{2\sin \frac{2x-4x}{2}\cos \frac{4x+2x}{2}} \\
& =\frac{-2\sin \frac{6x}{2}\sin \frac{2x}{2}}{2\sin \frac{-2x}{2}\cos \frac{6x}{2}} \\
& =\frac{\sin 3x}{\cos 3x} \\
& =\tan 3x
\end{align}$
Hence, verified.
