Use \(\sin(u\pm v)=\sin u\cos v\pm\cos u\sin v\):
\( \sin(\alpha+\beta)+\sin(\alpha-\beta)\) \(=(\sin\alpha\cos\beta+\cos\alpha\sin\beta)\) \(+(\sin\alpha\cos\beta-\cos\alpha\sin\beta)\) \(=2\sin\alpha\cos\beta. \)

Use \(\cos(u\pm v)=\cos u\cos v\mp\sin u\sin v\):
\( \cos(\alpha+\beta)+\cos(\alpha-\beta)\) \( =(\cos\alpha\cos\beta-\sin\alpha\sin\beta)\) \(+(\cos\alpha\cos\beta+\sin\alpha\sin\beta)\) \(=2\cos\alpha\cos\beta. \)

Recognize the sine–difference pattern \(\sin U\cos V-\cos U\sin V=\sin(U-V)\) with \(U=\alpha-\beta\), \(V=\alpha\):
\( \sin(\alpha-\beta)\cos\alpha-\cos(\alpha-\beta)\sin\alpha\) \(=\sin\big((\alpha-\beta)-\alpha\big)\) \(=\sin(-\beta)=-\sin\beta. \)

Start from the right-hand side and rewrite in sine/cosine:

\[ \frac{\csc\theta\,\csc\varphi}{\cot\theta+\cot\varphi} \;=\; \frac{\dfrac{1}{\sin\theta}\cdot\dfrac{1}{\sin\varphi}} {\dfrac{\cos\theta}{\sin\theta}+\dfrac{\cos\varphi}{\sin\varphi}} \]

Combine the denominator over a common denominator:

\[ =\; \frac{\dfrac{1}{\sin\theta\,\sin\varphi}} {\dfrac{\cos\theta\,\sin\varphi+\cos\varphi\,\sin\theta} {\sin\theta\,\sin\varphi}} \]

Invert and multiply:

\[ =\; \frac{1}{\cos\theta\,\sin\varphi+\cos\varphi\,\sin\theta} \]

Recognize the sine addition formula \(\sin(\theta+\varphi)=\sin\theta\cos\varphi+\cos\theta\sin\varphi\):

\[ =\; \frac{1}{\sin(\theta+\varphi)} \;=\; \csc(\theta+\varphi). \]

Hence \(\displaystyle \csc(\theta+\varphi)\equiv \dfrac{\csc\theta\,\csc\varphi}{\cot\theta+\cot\varphi}\).

Expand each cosine:

\[ \cos(5x+x) = \cos 5x\cos x - \sin 5x\sin x, \] \[ \cos(5x-x) = \cos 5x\cos x + \sin 5x\sin x. \]

Add the two results:

\[ \cos 6x + \cos 4x = (\cos 5x\cos x - \sin 5x\sin x) + (\cos 5x\cos x + \sin 5x\sin x) = 2\cos 5x\cos x. \]

Hence proved.