Question

9.7 notetaking with vocabulary (co

1. \\(\dfrac{2\sec^2 x - 2\tan^2 x}{\tan(-x)\cos(-x)}\\)

Explanation:

Step1: Use trigonometric identities

We know that \(\sec^{2}x = 1+\tan^{2}x\). So \(2\sec^{2}x-2\tan^{2}x=2(1 + \tan^{2}x)-2\tan^{2}x=2\). Also, \(\tan(-x)=-\tan x\) and \(\cos(-x)=\cos x\).
The original expression \(\frac{2\sec^{2}x - 2\tan^{2}x}{\tan(-x)\cos(-x)}\) becomes \(\frac{2}{(-\tan x)\cos x}\).

Step2: Simplify using \(\tan x=\frac{\sin x}{\cos x}\)

Substitute \(\tan x=\frac{\sin x}{\cos x}\) into \(\frac{2}{(-\tan x)\cos x}\), we get \(\frac{2}{-\frac{\sin x}{\cos x}\cdot\cos x}\).
Simplify the denominator \(-\frac{\sin x}{\cos x}\cdot\cos x=-\sin x\).
So the expression is \(\frac{2}{-\sin x}=- 2\csc x\).

Answer:

\(-2\csc x\)