Question

question 10 (mandatory) (1 point)
the identity that is not equivalent to sin x is
a) cos xtan x
b) \\( \frac { \sin x \tan x } { \sin ^ { 2 } x - \cos ^ { 2 } x } \\)
c) \\( \frac { \cos ^ { 2 } x \sin x } { ( 1 - \sin x ) ( 1 + \sin x ) } \\)
d) \\( \frac { 1 - \cos ^ { 2 } x } { \sin x } \\)

Explanation:

Step1: Simplify option a

Use the identity \(\tan x=\frac{\sin x}{\cos x}\).
\(\cos x\tan x=\cos x\times\frac{\sin x}{\cos x}=\sin x\)

Step2: Simplify option b

Use the identity \(\sin^{2}x-\cos^{2}x = -(\cos^{2}x-\sin^{2}x)=-(1 - 2\sin^{2}x)\) and \(\tan x=\frac{\sin x}{\cos x}\).
\(\frac{\sin x\tan x}{\sin^{2}x-\cos^{2}x}=\frac{\sin x\times\frac{\sin x}{\cos x}}{-(\cos^{2}x - \sin^{2}x)}\). This is not equal to \(\sin x\) in general.

Step3: Simplify option c

Use the identity \((1 - \sin x)(1+\sin x)=1-\sin^{2}x=\cos^{2}x\).
\(\frac{\cos^{2}x\sin x}{(1 - \sin x)(1+\sin x)}=\frac{\cos^{2}x\sin x}{\cos^{2}x}=\sin x\)

Step4: Simplify option d

Use the identity \(1-\cos^{2}x=\sin^{2}x\).
\(\frac{1-\cos^{2}x}{\sin x}=\frac{\sin^{2}x}{\sin x}=\sin x\)

Answer:

B. \(\frac{\sin x\tan x}{\sin^{2}x-\cos^{2}x}\)