Question

Pls hurry. thank you.

Answer 1

`\large\boxed{(1-2\cos^2\varphi)/(\sin\varphi\cos\varphi)-\tan\varph=-\cot\varphi}`

Explanation:

`(1-2\cos^2\varphi)/(\sin\varphi\cos\varphi)-\tan\varphi\qquad\text{use}\ \tan x=(\sin x)/(\cos x)\\\\=(1-\cos^2\varphi-\cos^2\varphi)/(\sin\varphi\cos\varphi)-(\sin\varphi)/(\cos\varphi)\qquad\text{use}\ \sin^2x+\cos^2x=1\to\sin^2x=1-\cos^2x\\\\=(\sin^2\varphi-\cos^2\varphi)/(\sin\varphi\cos\varphi)-(\sin\varphi\sin\varphi)/(\sin\varphi\cos\varphi)=(\sin^2\varphi-\cos^2\varphi-\sin^2\varphi)/(\sin\varphi\cos\varphi)`

`=((\sin^2\varphi-\sin^2\varphi)-\cos^2\varphi)/(\sin\varphi\cos\varphi)=(-\cos^2\varphi)/(\sin\varphi\cos\varphi)\qquad\text{cancel one}\ \cos\varphi\\\\=(-\cos\varphi)/(\sin\varphi)=-(\cos\varphi)/(\sin\varphi)\qquad\text{use}\ \cot x=(\cos x)/(\sin x)\\\\=\boxed{-\cot\varphi}`