Question

question 11 (mandatory) (1 point)
the identity that is not equivalent to \\( \sin x \\) is
a) \\( \cos x \tan x \\)
b) \\( \frac{\cos ^{2} x \sin x}{(1-\sin x)(1+\sin x)} \\)
c) \\( \frac{\sin x \tan x}{\sin ^{2} x-\cos ^{2} x} \\)
d) \\( \frac{1-\cos ^{2} x}{\sin x} \\)
question 12 (mandatory) (1 point)
which equation is the quotient identity?

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 difference - of - squares formula \((a - b)(a + b)=a^{2}-b^{2}\), so \((1-\sin x)(1 + \sin x)=1-\sin^{2}x\).
Then \(\frac{\cos^{2}x\sin x}{(1-\sin x)(1 + \sin x)}=\frac{\cos^{2}x\sin x}{1-\sin^{2}x}\).
Since \(1-\sin^{2}x=\cos^{2}x\), \(\frac{\cos^{2}x\sin x}{\cos^{2}x}=\sin x\)

Step3: Simplify option c

Use the identity \(\tan x=\frac{\sin x}{\cos x}\) and \(a^{2}-b^{2}=(a - b)(a + b)\), so \(\sin^{2}x-\cos^{2}x=-(\cos^{2}x-\sin^{2}x)\).
\(\frac{\sin x\tan x}{\sin^{2}x-\cos^{2}x}=\frac{\sin x\times\frac{\sin x}{\cos x}}{-(\cos^{2}x-\sin^{2}x)}=\frac{\frac{\sin^{2}x}{\cos x}}{-(\cos^{2}x-\sin^{2}x)}
eq\sin x\)

Step4: Simplify option d

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

Answer:

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