multiple choice\nidentify the choice that best completes the statement or answers the question.\n1. the…

multiple choice\nidentify the choice that best completes the statement or answers the question.\n1. the simplified form of \\( \\cos \\theta \\tan \\theta \\sin \\theta \\) is\na. \\( \\frac{\\sin ^{2} \\theta}{\\cos \\theta} \\)\nc. \\( \\sin ^{2} \\theta \\cos \\theta \\)\nb. \\( 1-\\cos ^{2} \\theta \\)\nd. \\( \\sin \\theta \\)\n2. using the definitions \\( \\sin \\theta=\\frac{y}{r}, \\cos \\theta=\\frac{x}{r} \\), and \\( \\tan \\theta=\\frac{y}{r} \\), the simplified form of the expression\n\\( \\frac{\\sin ^{2} \\theta+\\cos ^{2} \\theta}{\\frac{\\cos \\theta}{\\sin \\theta}} \\) is\na. \\( \\frac{x}{y} \\)\nc. \\( \\frac{x}{r} \\)\nb. \\( \\frac{y}{x} \\)\nd. \\( \\frac{y}{r} \\)\n3. the expression \\( \\frac{\\sin ^{2} \\theta+\\cos ^{2} \\theta}{\\frac{\\cos \\theta}{\\sin \\theta}} \\) is the equivalent to\na. \\( \\frac{\\cos \\theta}{\\sin \\theta} \\)\nc. \\( \\tan \\theta \\)\nb. \\( \\sin \\theta \\)\nd. \\( \\cos \\theta \\)\n4. the expression \\( \\tan \\theta\\left(\\frac{\\sin \\theta}{\\cos \\theta}\\right) \\) is equivalent to\na. \\( \\frac{x}{y} \\)\nc. \\( \\frac{y}{x} \\)\nb. \\( \\frac{x^{2}}{y^{2}} \\)\nd. \\( \\frac{y^{2}}{x^{2}} \\)\n5. the identity that is not equivalent to \\( \\cos x \\) is\na. \\( \\frac{\\sin x}{\\tan x} \\)\nc. \\( \\frac{\\tan x\\left(1-\\sin ^{2} x\\right)}{\\cos x \\sin x} \\)\nb. \\( \\frac{1-\\sin ^{2} x}{\\cos x} \\)\nd. \\( \\frac{\\sin ^{2} x \\cos x}{1-\\cos ^{2} x} \\)

multiple choice\nidentify the choice that best completes the statement or answers the question.\n1. the simplified form of \\( \\cos \\theta \\tan \\theta \\sin \\theta \\) is\na. \\( \\frac{\\sin ^{2} \\theta}{\\cos \\theta} \\)\nc. \\( \\sin ^{2} \\theta \\cos \\theta \\)\nb. \\( 1-\\cos ^{2} \\theta \\)\nd. \\( \\sin \\theta \\)\n2. using the definitions \\( \\sin \\theta=\\frac{y}{r}, \\cos \\theta=\\frac{x}{r} \\), and \\( \\tan \\theta=\\frac{y}{r} \\), the simplified form of the expression\n\\( \\frac{\\sin ^{2} \\theta+\\cos ^{2} \\theta}{\\frac{\\cos \\theta}{\\sin \\theta}} \\) is\na. \\( \\frac{x}{y} \\)\nc. \\( \\frac{x}{r} \\)\nb. \\( \\frac{y}{x} \\)\nd. \\( \\frac{y}{r} \\)\n3. the expression \\( \\frac{\\sin ^{2} \\theta+\\cos ^{2} \\theta}{\\frac{\\cos \\theta}{\\sin \\theta}} \\) is the equivalent to\na. \\( \\frac{\\cos \\theta}{\\sin \\theta} \\)\nc. \\( \\tan \\theta \\)\nb. \\( \\sin \\theta \\)\nd. \\( \\cos \\theta \\)\n4. the expression \\( \\tan \\theta\\left(\\frac{\\sin \\theta}{\\cos \\theta}\\right) \\) is equivalent to\na. \\( \\frac{x}{y} \\)\nc. \\( \\frac{y}{x} \\)\nb. \\( \\frac{x^{2}}{y^{2}} \\)\nd. \\( \\frac{y^{2}}{x^{2}} \\)\n5. the identity that is not equivalent to \\( \\cos x \\) is\na. \\( \\frac{\\sin x}{\\tan x} \\)\nc. \\( \\frac{\\tan x\\left(1-\\sin ^{2} x\\right)}{\\cos x \\sin x} \\)\nb. \\( \\frac{1-\\sin ^{2} x}{\\cos x} \\)\nd. \\( \\frac{\\sin ^{2} x \\cos x}{1-\\cos ^{2} x} \\)

Answer

Explanation:

Step1: Simplify (\cos\theta\tan\theta\sin\theta)

Recall that (\tan\theta=\frac{\sin\theta}{\cos\theta}). Then (\cos\theta\tan\theta\sin\theta=\cos\theta\times\frac{\sin\theta}{\cos\theta}\times\sin\theta=\sin^{2}\theta). Also, since (\sin^{2}\theta = 1-\cos^{2}\theta) (using the Pythagorean identity (\sin^{2}\theta+\cos^{2}\theta = 1)).

Step2: Simplify (\frac{\sin^{2}\theta+\cos^{2}\theta}{\frac{\cos\theta}{\sin\theta}})

First, use the Pythagorean identity (\sin^{2}\theta+\cos^{2}\theta = 1). Then the expression becomes (\frac{1}{\frac{\cos\theta}{\sin\theta}}=\frac{\sin\theta}{\cos\theta}=\tan\theta). But if we use the definitions (\sin\theta=\frac{y}{r},\cos\theta=\frac{x}{r}), (\frac{\sin^{2}\theta+\cos^{2}\theta}{\frac{\cos\theta}{\sin\theta}}=\frac{1}{\frac{\frac{x}{r}}{\frac{y}{r}}}=\frac{y}{x})

Step3: Simplify (\tan\theta(\frac{\sin\theta}{\cos\theta}))

Since (\tan\theta=\frac{\sin\theta}{\cos\theta}), then (\tan\theta(\frac{\sin\theta}{\cos\theta})=\frac{\sin\theta}{\cos\theta}\times\frac{\sin\theta}{\cos\theta}=\frac{\sin^{2}\theta}{\cos^{2}\theta}). Using (\sin\theta=\frac{y}{r},\cos\theta=\frac{x}{r}), we get (\frac{y^{2}}{x^{2}})

Step4: Simplify each option for the non - equivalent to (\cos x)

  • For option a: (\frac{\sin x}{\tan x}=\frac{\sin x}{\frac{\sin x}{\cos x}}=\cos x)
  • For option b: (\frac{1 - \sin^{2}x}{\cos x}=\frac{\cos^{2}x}{\cos x}=\cos x)
  • For option c: (\frac{\tan x(1-\sin^{2}x)}{\cos x\sin x}=\frac{\frac{\sin x}{\cos x}\times\cos^{2}x}{\cos x\sin x}=1)
  • For option d: (\frac{\sin^{2}x\cos x}{1 - \cos^{2}x}=\frac{\sin^{2}x\cos x}{\sin^{2}x}=\cos x)

Answer:

  1. B. (1-\cos^{2}\theta)
  2. B. (\frac{y}{x})
  3. C. (\tan\theta)
  4. D. (\frac{y^{2}}{x^{2}})
  5. C. (\frac{\tan x(1-\sin^{2}x)}{\cos x\sin x})