Answer:
[tex]\large\boxed{\dfrac{1-2\cos^2\varphi}{\sin\varphi\cos\varphi}-\tan\varph=-\cot\varphi}[/tex]
Step-by-step explanation:
[tex]\dfrac{1-2\cos^2\varphi}{\sin\varphi\cos\varphi}-\tan\varphi\qquad\text{use}\ \tan x=\dfrac{\sin x}{\cos x}\\\\=\dfrac{1-\cos^2\varphi-\cos^2\varphi}{\sin\varphi\cos\varphi}-\dfrac{\sin\varphi}{\cos\varphi}\qquad\text{use}\ \sin^2x+\cos^2x=1\to\sin^2x=1-\cos^2x\\\\=\dfrac{\sin^2\varphi-\cos^2\varphi}{\sin\varphi\cos\varphi}-\dfrac{\sin\varphi\sin\varphi}{\sin\varphi\cos\varphi}=\dfrac{\sin^2\varphi-\cos^2\varphi-\sin^2\varphi}{\sin\varphi\cos\varphi}[/tex]
[tex]=\dfrac{(\sin^2\varphi-\sin^2\varphi)-\cos^2\varphi}{\sin\varphi\cos\varphi}=\dfrac{-\cos^2\varphi}{\sin\varphi\cos\varphi}\qquad\text{cancel one}\ \cos\varphi\\\\=\dfrac{-\cos\varphi}{\sin\varphi}=-\dfrac{\cos\varphi}{\sin\varphi}\qquad\text{use}\ \cot x=\dfrac{\cos x}{\sin x}\\\\=\boxed{-\cot\varphi}[/tex]
