Question

1. the identity that is not equivalent to \\( \sin x \\) is

a. \\( \frac { 1 - \cos ^ { 2 } x } { \sin x } \\)
c. \\( \frac { \cos ^ { 2 } x \sin x } { ( 1 - \sin x ) ( 1 + \sin x ) } \\)
b. \\( \cos x \tan x \\)
d. \\( \frac { \sin x \tan x } { \sin ^ { 2 } x - \cos ^ { 2 } x } \\)

1. the identity that is equivalent to \\( \tan x \\) is

a. \\( \frac { \cos x } { \sin x } \\)
c. \\( \frac { \cos x \tan x } { 1 - \sin ^ { 2 } x } \\)
b. \\( \frac { \cos x \tan x \left( 1 - \sin ^ { 2 } x \
ight) } { \sin ^ { 2 } x + \cos ^ { 2 } x } \\)
d. \\( \frac { 1 - \cos ^ { 2 } x } { \sin x \cos x } \\)

1. the factored form of \\( \cos ^ { 2 } x - 2 \cos x + 1 \\) is

a. \\( ( \cos x - 1 ) ( \cos x - 1 ) \\)
c. \\( \cos x ( \cos x - 2 ) + 1 \\)
b. \\( ( \cos x - 1 ) ( \cos x + 1 ) \\)
d. 1

1. expanding the expression \\( ( 1 - \cos x ) ( 1 + \cos x ) \\) gives

a. \\( \tan x \\)
c. \\( 1 + \cos ^ { 2 } x \\)
b. \\( \sin x \\)
d. \\( \sin ^ { 2 } x \\)

1. the simplified form of the expression \\( \frac { \sin x \sin x } { ( 1 - \sin x ) ( 1 + \sin x ) } \\) is

a. \\( \frac { \sin ^ { 2 } x } { \cos x } \\)
c. \\( \tan ^ { 2 } x \\)
b. \\( \frac { \sin ^ { 2 } x } { \sin x } \\)
d. \\( \frac { \sin ^ { 2 } x } { 1 + \sin ^ { 2 } x } \\)

1. the simplified form of \\( \frac { \cos \theta \sin \theta } { 1 - \cos ^ { 2 } \theta } \\) is

a. \\( \frac { 1 } { \tan \theta } \\)
c. \\( \frac { 1 } { \cos \theta } \\)
b. \\( \frac { 1 } { \sin \theta } \\)
d. \\( \tan \theta \\)

1. the expression that is equivalent to \\( 1 - 2 \sin \theta + \sin ^ { 2 } \theta \\) is

a. \\( ( 1 - \cos \theta ) ^ { 2 } \\)
c. \\( 2 - \sin ^ { 2 } \theta \\)
b. \\( ( 1 - \sin \theta ) ^ { 2 } \\)
d. \\( 2 \sin \theta - 1 \\)

Explanation:

Step1: Simplify option a

$$\begin{align*} \frac{1-\cos^{2}x}{\sin x}&=\frac{\sin^{2}x}{\sin x}\\ &=\sin x \end{align*}$$

Step2: Simplify option b

$$\begin{align*} \cos x\tan x&=\cos x\times\frac{\sin x}{\cos x}\\ &=\sin x \end{align*}$$

Step3: Simplify option c

$$\begin{align*} \frac{\cos^{2}x\sin x}{(1 - \sin x)(1+\sin x)}&=\frac{\cos^{2}x\sin x}{1-\sin^{2}x}\\ &=\frac{\cos^{2}x\sin x}{\cos^{2}x}\\ &=\sin x \end{align*}$$

Step4: Simplify option d

$$\begin{align*} \frac{\sin x\tan x}{\sin^{2}x-\cos^{2}x}&=\frac{\sin x\times\frac{\sin x}{\cos x}}{\sin^{2}x-\cos^{2}x}\\ &=\frac{\frac{\sin^{2}x}{\cos x}}{\sin^{2}x-\cos^{2}x}\\ & eq\sin x \end{align*}$$

Answer:

d. $\frac{\sin x\tan x}{\sin^{2}x-\cos^{2}x}$