Question

question 4 (mandatory) (1 point)
the identity that is equivalent to tan x is
a) \\( \frac { \cos x } { \sin x } \\)
b) \\( \frac { \cos x \tan x ( 1 - \sin ^ { 2 } x ) } { \sin ^ { 2 } x + \cos ^ { 2 } x } \\)
c) \\( \frac { 1 - \cos ^ { 2 } x } { \sin x \cos x } \\)
d) \\( \frac { \cos x \tan x } { 1 - \sin ^ { 2 } x } \\)

Explanation:

Step1: Recall trigonometric identities

We know that \(\tan x=\frac{\sin x}{\cos x}\), \(\sin^{2}x+\cos^{2}x = 1\) (so \(1-\sin^{2}x=\cos^{2}x\) and \(1 - \cos^{2}x=\sin^{2}x\))

Step2: Simplify option b

Simplify \(\frac{\cos x\tan x(1-\sin^{2}x)}{\sin^{2}x+\cos^{2}x}\)
Since \(\sin^{2}x+\cos^{2}x = 1\) and \(1-\sin^{2}x=\cos^{2}x\), the expression becomes \(\cos x\tan x\cos^{2}x\)
Substitute \(\tan x=\frac{\sin x}{\cos x}\), we get \(\cos x\times\frac{\sin x}{\cos x}\times\cos^{2}x=\sin x\cos^{2}x
eq\tan x\)

Step3: Simplify option c

Simplify \(\frac{1-\cos^{2}x}{\sin x\cos x}\)
Since \(1 - \cos^{2}x=\sin^{2}x\), the expression becomes \(\frac{\sin^{2}x}{\sin x\cos x}=\frac{\sin x}{\cos x}=\tan x\)

Step4: Simplify option a

\(\frac{\cos x}{\sin x}=\cot x
eq\tan x\)

Step5: Simplify option d

Simplify \(\frac{\cos x\tan x}{1-\sin^{2}x}\)
Since \(1-\sin^{2}x=\cos^{2}x\) and \(\tan x=\frac{\sin x}{\cos x}\), the expression becomes \(\frac{\cos x\times\frac{\sin x}{\cos x}}{\cos^{2}x}=\frac{\sin x}{\cos^{2}x}
eq\tan x\)

Answer:

C. \(\frac{1 - \cos^{2}x}{\sin x\cos x}\)