Question

Factor; then use fundamental identities to simplify the expression below and determine which of the following is NOT equivalent tan^3x-tan^2x+tanx-1a.sec^2x(sinx-cosx/cosx)b.sinx-cosx/(cos^3x)c.sec^2x-tan^2xd. (sinx/cos^3x)-sec^2xe.(tanx)(sec^2x)-sec^2x

Answer 1

We have the following expression

$\tan ^3x-\tan ^2x+\tan x-1$

Factoring

$\begin{gathered} \tan ^2x(\tan x-1)+\tan x-1 \\ (\tan x-1)(\tan ^2x+1) \end{gathered}$

Now, we need to compare every option

For option E = Equivalent

$\begin{gathered} (\tan x-1)(\tan ^2x+1) \\ (\tan x-1)\cdot(\sec ^2x) \\ \tan x\cdot\sec ^2x-\sec ^2x \end{gathered}$

For option D = Equivalent

$\begin{gathered} \tan x\cdot\sec ^2x-\sec ^2x \\ \frac{\sin x}{\cos x}\cdot\frac{1}{\cos^2x}-\sec ^2x \\ \frac{\sin x}{\cos^3x}-\sec ^2x \end{gathered}$

For option C = Not Equivalent

$\begin{gathered} \tan x\cdot\sec ^2x-\sec ^2x \\ \text{ we can not derive any expression like option C} \end{gathered}$

For option B = Equivalent

$\begin{gathered} \tan x\cdot\sec ^2x-\sec ^2x \\ \frac{\sin x}{\cos x}\cdot\frac{1}{\cos^2x}-\frac{1}{\cos^2x} \\ \frac{\sin x}{\cos^3x}-\frac{\cos x}{\cos^3x} \\ \frac{\sin x-\cos x}{\cos^3x} \end{gathered}$

For option A = Equivalent

$\begin{gathered} \frac{\sin x-\cos x}{\cos^3x} \\ \frac{1}{\cos^2x}\cdot\frac{\sin x-\cos x}{\cos ^{}x} \\ \sec ^2x\cdot\frac{\sin x-\cos x}{\cos ^{}x} \end{gathered}$