Proving Trigonometric Identities

Question 1 of 6

Identify if the equation is correct

\frac{\cos θ}{1 - \sin θ} = \sec θ + \tan θ

$(cos theta)/(1-sin theta)=sec theta+tan theta$

1. $Correct$ $\text(Correct)$
2. $Incorrect$ $\text(Incorrect)$

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Trigonometric Identities

\sec θ = \frac{1}{cos θ}

$\sec\theta=\frac{1}{\text{cos}\theta}$

\csc θ = \frac{1}{sin θ}

$\csc\theta=\frac{1}{\text{sin}\theta}$

\cot θ = \frac{1}{tan θ}

$\cot\theta=\frac{1}{\text{tan}\theta}$

\tan θ = \frac{sin θ}{cos θ}

$\tan\theta=\frac{\text{sin}\theta}{\text{cos}\theta}$

\cot θ = \frac{cos θ}{sin θ}

$\cot\theta=\frac{\text{cos}\theta}{\text{sin}\theta}$

Work on changing the left side of the equation to make it equal to the value of the right side

	$\frac{cos θ}{1 - sin θ}$ $(\text(cos) theta)/(1-\text(sin) theta)$

$=$ $=$	$\frac{\text{cos}\;\theta\cdot(\color{#CC0000}{1+\text{sin}\;\theta})}{1-\text{sin}\;\theta\cdot(\color{#CC0000}{1+\text{sin}\;\theta})}$	Multiply the numerator and denominator by $1 + sin θ$ $1+\text(sin) theta$

$=$ $=$	$\frac{cos θ (1 + sin θ)}{1 - {sin}^{2} θ}$ $(\text(cos) theta(1+\text(sin) theta))/(1-\text(sin)^2 theta)$	Difference of two squares

Recall that

{sin}^{2} θ + {cos}^{2} θ = 1

$\text(sin)^2theta+\text(cos)^2theta=1$ . Derive this formula to further simplify the expression

${sin}^{2} θ + {cos}^{2} θ$ $\text(sin)^2theta+\text(cos)^2theta$	$=$ $=$	$1$ $1$
${sin}^{2} θ + {cos}^{2} θ$ $\text(sin)^2theta+\text(cos)^2theta$ $- {sin}^{2} θ$ $-\text(sin)^2theta$	$=$ $=$	$1$ $1$ $- {sin}^{2} θ$ $-\text(sin)^2theta$	Subtract ${sin}^{2} θ$ $\text(sin)^2theta$ from both sides
${cos}^{2} θ$ $\text(cos)^2theta$	$=$ $=$	$1 - {sin}^{2} θ$ $1-\text(sin)^2theta$

	$\frac{cos θ (1 + sin θ)}{1 - {sin}^{2} θ}$ $(\text(cos) theta(1+\text(sin) theta))/(1-\text(sin)^2 theta)$

$=$ $=$	$\frac{cos θ (1 + sin θ)}{{cos}^{2} θ}$ $(\text(cos) theta(1+\text(sin) theta))/(\text(cos)^2theta)$	$1 - {sin}^{2} θ = {cos}^{2} θ$ $1-\text(sin)^2theta=\text(cos)^2theta$

$=$ $=$	$\frac{1 + sin θ}{cos θ}$ $(1+\text(sin) theta)/(\text(cos) theta)$	$\frac{cos θ}{{cos}^{2} θ} = \frac{1}{cos θ}$ $(\text(cos) theta)/(\text(cos)^2theta)=1/(\text(cos) theta)$

$=$ $=$	$\frac{1}{cos θ} + \frac{sin θ}{cos θ}$ $1/(\text(cos) theta)+(\text(sin) theta)/(\text(cos)theta)$	Separate the values of the numerator

$=$ $=$	$sec θ + \frac{sin θ}{cos θ}$ $\text(sec)theta+(\text(sin) theta)/(\text(cos)theta)$	$\frac{1}{cos θ} = sec θ$ $1/(\text(cos) theta)=\text(sec)theta$

$=$ $=$	$sec θ + tan θ$ $\text(sec)theta+\text(tan)theta$	$\frac{sin θ}{cos θ} = tan θ$ $(\text(sin) theta)/(\text(cos)theta)=\text(tan)theta$

This means that

\frac{cos θ}{1 - sin θ} = sec θ + tan θ

$(\text(cos) theta)/(1-\text(sin) theta)=\text(sec) theta+\text(tan) theta$

Correct

$\text(Correct)$

Question 2 of 6

Identify if the equation is correct

\frac{\sin A}{1 + \cos A} + \frac{1 + \cos A}{\sin A} = 2 \csc A

$(sin A)/(1+cos A)+(1+cos A)/(sin A)=2 csc A$

1. $Correct$ $\text(Correct)$
2. $Incorrect$ $\text(Incorrect)$

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Trigonometric Identities

\sec θ = \frac{1}{cos θ}

$\sec\theta=\frac{1}{\text{cos}\theta}$

\csc θ = \frac{1}{sin θ}

$\csc\theta=\frac{1}{\text{sin}\theta}$

\cot θ = \frac{1}{tan θ}

$\cot\theta=\frac{1}{\text{tan}\theta}$

\tan θ = \frac{sin θ}{cos θ}

$\tan\theta=\frac{\text{sin}\theta}{\text{cos}\theta}$

\cot θ = \frac{cos θ}{sin θ}

$\cot\theta=\frac{\text{cos}\theta}{\text{sin}\theta}$

Work on changing the left side of the equation to make it equal to the value of the right side

	$\frac{sin A}{1 + cos A} + \frac{1 + cos A}{sin A}$ $(\text(sin) A)/(1+\text(cos) A)+(1+\text(cos) A)/(\text(sin) A)$

$=$ $=$	$\frac{{sin}^{2} A + {(1 + cos A)}^{2}}{(1 + cos A) sin A}$ $(\text(sin)^2A+(1+\text(cos) A)^2)/((1+\text(cos) A)\text(sin) A)$	Apply the rule of adding fractions

$=$ $=$	$\frac{{sin}^{2} A + (1 + cos A) (1 + cos A)}{(1 + cos A) sin A}$ $(\text(sin)^2A+(1+\text(cos) A)(1+\text(cos) A))/((1+\text(cos) A)\text(sin) A)$	Expand

$=$ $=$	$\frac{{sin}^{2} A + 1 + 2 cos A + {cos}^{2} A}{(1 + cos A) sin A}$ $(\text(sin)^2A+1+2\text(cos) A+\text(cos)^2A)/((1+\text(cos) A)\text(sin) A)$

Recall that

{sin}^{2} θ + {cos}^{2} θ = 1

$\text(sin)^2theta+\text(cos)^2theta=1$ . Derive this formula to further simplify the expression

	$\frac{{sin}^{2} A + 1 + 2 cos A + {cos}^{2} A}{(1 + cos A) sin A}$ $(\text(sin)^2A+1+2\text(cos) A+\text(cos)^2A)/((1+\text(cos) A)\text(sin) A)$

$=$ $=$	$\frac{1 + 1 + 2 cos A}{(1 + cos A) sin A}$ $(1+1+2\text(cos) A)/((1+\text(cos) A)\text(sin) A)$	${sin}^{2} A + {cos}^{2} A = 1$ $\text(sin)^2A+\text(cos)^2A=1$

$=$ $=$	$\frac{2 + 2 cos A}{(1 + cos A) sin A}$ $(2+2\text(cos) A)/((1+\text(cos) A)\text(sin) A)$	Combine like terms

$=$ $=$	$\frac{2 (1 + cos A)}{(1 + cos A) sin A}$ $(2(1+\text(cos) A))/((1+\text(cos) A)\text(sin) A)$	Factor out $2$ $2$

$=$ $=$	$\frac{2}{sin A}$ $2/(\text(sin) A)$	$\frac{1 + cos A}{1 + cos A} = 1$ $(1+\text(cos) A)/(1+\text(cos) A)=1$

$=$ $=$	$2 csc A$ $2\text(csc) A$	$\frac{1}{sin A} = csc A$ $1/(\text(sin) A)=\text(csc) A$

This means that

\frac{\sin A}{1 + \cos A} + \frac{1 + \cos A}{\sin A} = 2 \csc A

$(sin A)/(1+cos A)+(1+cos A)/(sin A)=2 csc A$

Correct

$\text(Correct)$

Question 3 of 6

Identify if the equation is correct

\frac{1 + \cot θ}{\csc θ} - \frac{\sec θ}{\tan θ + \cot θ} = \cos θ

$(1+cot theta)/(csc theta)-(sec theta)/(tan theta+cot theta)=cos theta$

1. $Correct$ $\text(Correct)$
2. $Incorrect$ $\text(Incorrect)$

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Trigonometric Identities

\sec θ = \frac{1}{cos θ}

$\sec\theta=\frac{1}{\text{cos}\theta}$

\csc θ = \frac{1}{sin θ}

$\csc\theta=\frac{1}{\text{sin}\theta}$

\cot θ = \frac{1}{tan θ}

$\cot\theta=\frac{1}{\text{tan}\theta}$

\tan θ = \frac{sin θ}{cos θ}

$\tan\theta=\frac{\text{sin}\theta}{\text{cos}\theta}$

\cot θ = \frac{cos θ}{sin θ}

$\cot\theta=\frac{\text{cos}\theta}{\text{sin}\theta}$

Work on changing the left side of the equation to make it equal to the value of the right side

First term

	$\frac{1 + cot θ}{csc θ}$ $(1+\text(cot) theta)/(\text(csc) theta)$

$=$ $=$	$\frac{1}{csc θ} (1 + cot θ)$ $1/(\text(csc) theta)(1+\text(cot) theta)$	Factorise

$=$ $=$	$sin θ (1 + cot θ)$ $\text(sin) theta(1+\text(cot) theta)$	$\frac{1}{csc θ} = sin θ$ $1/(\text(csc) theta)=\text(sin) theta$

$=$ $=$	$sin θ (1 + \frac{cos θ}{sin θ})$ $\text(sin) theta(1+(\text(cos) theta)/(\text(sin) theta))$	$cot θ = \frac{cos θ}{sin θ}$ $\text(cot) theta=(\text(cos) theta)/(\text(sin) theta)$

$=$ $=$	$sin θ + \frac{sin θ \cdot cos θ}{sin θ}$ $\text(sin) theta+(\text(sin) theta*\text(cos) theta)/(\text(sin) theta)$	Expand

$=$ $=$	$sin θ + cos θ$ $\text(sin) theta+\text(cos) theta$	$\frac{sin θ}{sin θ} = 1$ $(\text(sin) theta)/(\text(sin) theta)=1$

Second term

	$\frac{sec θ}{tan θ + cot θ}$ $(\text(sec) theta)/(\text(tan) theta+\text(cot) theta)$

$=$ $=$	$\frac{\frac{1}{cos θ}}{tan θ + cot θ}$ $(1/(\text(cos) theta))/(\text(tan) theta+\text(cot) theta)$	$sec θ = \frac{1}{cos θ}$ $\text(sec) theta=1/(\text(cos) theta)$

$=$ $=$	$\frac{\frac{1}{cos θ}}{\frac{sin θ}{cos θ} + cot θ}$ $(1/(\text(cos) theta))/((\text(sin) theta)/(\text(cos) theta)+\text(cot) theta)$	$tan θ = \frac{sin θ}{cos θ}$ $\text(tan) theta=(\text(sin) theta)/(\text(cos) theta)$

$=$ $=$	$\frac{\frac{1}{cos θ}}{\frac{sin θ}{cos θ} + \frac{cos θ}{sin θ}}$ $(1/(\text(cos) theta))/((\text(sin) theta)/(\text(cos) theta)+(\text(cos) theta)/(\text(sin) theta)$	$cot θ = \frac{cos θ}{sin θ}$ $\text(cot) theta=(\text(cos) theta)/(\text(sin) theta)$

$=$ $=$	$\frac{\frac{1}{cos θ}}{\frac{{sin}^{2} θ + {cos}^{2} θ}{cos θ sin θ}}$ $(1/(\text(cos) theta))/((\text(sin)^2theta+\text(cos)^2theta)/(\text(cos) theta \text(sin) theta)$	Apply rule of adding fractions to the denominator

$=$ $=$	$\frac{\frac{1}{cos θ}}{\frac{1}{cos θ sin θ}}$ $(1/(\text(cos) theta))/(1/(\text(cos) theta \text(sin) theta)$	${sin}^{2} θ + {cos}^{2} θ = 1$ $\text(sin)^2theta+\text(cos)^2theta=1$

$=$ $=$	$\frac{1}{cos θ} \cdot \frac{cos θ sin θ}{1}$ $1/(\text(cos) theta)*(\text(cos) theta \text(sin) theta)/1$	Apply rule of dividing fractions

$=$ $=$	$sin θ$ $\text(sin) theta$

Finally, combine the two terms and subtract

		$sin θ + cos θ$ $\text(sin) theta+\text(cos) theta$ $-$ $-$ $sin θ$ $\text(sin) theta$
	$=$ $=$	$cos θ$ $\text(cos) theta$

This means that

\frac{1 + cot θ}{csc θ} - \frac{sec θ}{tan θ + cot θ} = cos θ

$(1+\text(cot) theta)/(\text(csc) theta)-(\text(sec) theta)/(\text(tan) theta+\text(cot) theta)=\text(cos) theta$

Correct

$\text(Correct)$

Question 4 of 6

Identify if the equation is correct

\tan θ (1 - \cot^{2} θ) + \cot θ (1 - \tan^{2} θ) = 0

$tan theta(1-cot^2theta)+cot theta(1-tan^2theta)=0$

1. $Incorrect$ $\text(Incorrect)$
2. $Correct$ $\text(Correct)$

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Trigonometric Identities

\sec θ = \frac{1}{cos θ}

$\sec\theta=\frac{1}{\text{cos}\theta}$

\csc θ = \frac{1}{sin θ}

$\csc\theta=\frac{1}{\text{sin}\theta}$

\cot θ = \frac{1}{tan θ}

$\cot\theta=\frac{1}{\text{tan}\theta}$

\tan θ = \frac{sin θ}{cos θ}

$\tan\theta=\frac{\text{sin}\theta}{\text{cos}\theta}$

\cot θ = \frac{cos θ}{sin θ}

$\cot\theta=\frac{\text{cos}\theta}{\text{sin}\theta}$

Work on changing the left side of the equation to make it equal to the value of the right side

First term

	$tan θ (1 - {cot}^{2} θ)$ $\text(tan) theta(1-\text(cot)^2theta)$

$=$ $=$	$tan θ - (tan θ \cdot {cot}^{2} θ)$ $\text(tan) theta-(\text(tan) theta*\text(cot)^2theta)$	Expand

$=$ $=$	$tan θ - (\frac{sin θ}{cos θ} \cdot {cot}^{2} θ)$ $\text(tan) theta-((\text(sin) theta)/(\text(cos) theta)*\text(cot)^2theta)$	$tan θ = \frac{sin θ}{cos θ}$ $\text(tan) theta=(\text(sin) theta)/(\text(cos) theta)$

$=$ $=$	$tan θ - (\frac{sin θ}{cos θ} \cdot \frac{{cos}^{2} θ}{{sin}^{2} θ})$ $\text(tan) theta-((\text(sin) theta)/(\text(cos) theta)*(\text(cos)^2theta)/(\text(sin)^2theta))$	${cot}^{2} θ = \frac{{cos}^{2} θ}{{sin}^{2} θ}$ $\text(cot)^2theta=(\text(cos)^2theta)/(\text(sin)^2theta)$

$=$ $=$	$tan θ - (\frac{cos θ}{sin θ})$ $\text(tan) theta-((\text(cos) theta)/(\text(sin) theta))$	Simplify

$=$ $=$	$tan θ - cot θ$ $\text(tan) theta-\text(cot) theta$	$\frac{cos θ}{sin θ} = cot θ$ $(\text(cos) theta)/(\text(sin) theta)=\text(cot) theta$

Second term

	$cot θ (1 - {tan}^{2} θ)$ $\text(cot) theta(1-\text(tan)^2theta)$

$=$ $=$	$cot θ - (cot θ \cdot {tan}^{2} θ)$ $\text(cot) theta-(\text(cot) theta*\text(tan)^2theta)$	Expand

$=$ $=$	$cot θ - (\frac{cos θ}{sin θ} \cdot {tan}^{2} θ)$ $\text(cot) theta-((\text(cos) theta)/(\text(sin) theta)*\text(tan)^2theta)$	$cot θ = \frac{cos θ}{sin θ}$ $\text(cot) theta=(\text(cos) theta)/(\text(sin) theta)$

$=$ $=$	$cot θ - (\frac{cos θ}{sin θ} \cdot \frac{{sin}^{2} θ}{{cos}^{2} θ})$ $\text(cot) theta-((\text(cos) theta)/(\text(sin) theta)*(\text(sin)^2theta)/(\text(cos)^2theta))$	${tan}^{2} θ = \frac{{sin}^{2} θ}{{cos}^{2} θ}$ $\text(tan)^2theta=(\text(sin)^2theta)/(\text(cos)^2theta)$

$=$ $=$	$cot θ - (\frac{sin θ}{cos θ})$ $\text(cot) theta-((\text(sin) theta)/(\text(cos) theta))$	Simplify

$=$ $=$	$cot θ - tan θ$ $\text(cot) theta-\text(tan) theta$	$\frac{sin θ}{cos θ} = tan θ$ $(\text(sin) theta)/(\text(cos) theta)=\text(tan) theta$

Finally, combine the two terms and add

		$tan θ - cot θ$ $\text(tan) theta-\text(cot) theta$ $+$ $+$ $cot θ - tan θ$ $\text(cot) theta-\text(tan) theta$
	$=$ $=$	$0$ $0$

This means that

tan θ (1 - {cot}^{2} θ) + cot θ (1 - {tan}^{2} θ) = 0

$\text(tan) theta(1-\text(cot)^2theta)+\text(cot) theta(1-\text(tan)^2theta)=0$

Correct

$\text(Correct)$

Question 5 of 6

Identify if the equation is correct

\csc^{4} A - \cot^{4} A = \frac{1 + \cos^{2} A}{\sin^{2} A}

$csc^4A-cot^4A=(1+cos^2A)/(sin^2A)$

1. $Correct$ $\text(Correct)$
2. $Incorrect$ $\text(Incorrect)$

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Trigonometric Identities

\sec θ = \frac{1}{cos θ}

$\sec\theta=\frac{1}{\text{cos}\theta}$

\csc θ = \frac{1}{sin θ}

$\csc\theta=\frac{1}{\text{sin}\theta}$

\cot θ = \frac{1}{tan θ}

$\cot\theta=\frac{1}{\text{tan}\theta}$

\tan θ = \frac{sin θ}{cos θ}

$\tan\theta=\frac{\text{sin}\theta}{\text{cos}\theta}$

\cot θ = \frac{cos θ}{sin θ}

$\cot\theta=\frac{\text{cos}\theta}{\text{sin}\theta}$

Work on changing the left side of the equation to make it equal to the value of the right side

	${csc}^{4} A - {cot}^{4} A$ $\text(csc)^4A-\text(cot)^4A$
$=$ $=$	$({csc}^{2} A - {cot}^{2} A) ({csc}^{2} A + {cot}^{2} A)$ $(\text(csc)^2A-\text(cot)^2A)(\text(csc)^2A+\text(cot)^2A)$	Factorise
$=$ $=$	$[{csc}^{2} A - ({csc}^{2} A - 1)] [{csc}^{2} A + ({csc}^{2} A - 1)]$ $[\text(csc)^2A-(\text(csc)^2A-1)][\text(csc)^2A+(\text(csc)^2A-1)]$	${cot}^{2} A = {csc}^{2} A - 1$ $\text(cot)^2A=\text(csc)^2A-1$
$=$ $=$	$({csc}^{2} A - {csc}^{2} A + 1) ({csc}^{2} A + {csc}^{2} A - 1)$ $(\text(csc)^2A-\text(csc)^2A+1)(\text(csc)^2A+\text(csc)^2A-1)$	Adjust the subtracting values
$=$ $=$	$(1) (2 {csc}^{2} A - 1)$ $(1)(2\text(csc)^2A-1)$	Combine like terms

$=$ $=$	$(1) (2 \cdot \frac{1}{{sin}^{2} A} - 1)$ $(1)(2*1/(\text(sin)^2A)-1)$	${csc}^{2} A = \frac{1}{{sin}^{2} A}$ $\text(csc)^2A=1/(\text(sin)^2A)$

$=$ $=$	$\frac{2 - {sin}^{2} A}{{sin}^{2} A}$ $(2-\text(sin)^2A)/(\text(sin)^2A)$	Apply rule of subtracting fractions

$=$ $=$	$\frac{2 - (1 - {cos}^{2} A)}{{sin}^{2} A}$ $(2-(1-\text(cos)^2A))/(\text(sin)^2A)$	${sin}^{2} A = 1 - {cos}^{2} A$ $\text(sin)^2A=1-\text(cos)^2A$

$=$ $=$	$\frac{2 - 1 + {cos}^{2} A}{{sin}^{2} A}$ $(2-1+\text(cos)^2A)/(\text(sin)^2A)$	Adjust the subtracting values

$=$ $=$	$\frac{1 + {cos}^{2} A}{{sin}^{2} A}$ $(1+\text(cos)^2A)/(\text(sin)^2A)$

This means that

{csc}^{4} A - {cot}^{4} A = \frac{1 + {cos}^{2} A}{{sin}^{2} A}

$\text(csc)^4A-\text(cot)^4A=(1+\text(cos)^2A)/(\text(sin)^2A)$

Correct

$\text(Correct)$

Question 6 of 6

Identify if the equation is correct

\frac{\sin^{2} θ - \tan^{2} θ}{1 - \sec^{2} θ} = \sin^{2} θ

$(sin^2theta-tan^2theta)/(1-sec^2theta)=sin^2theta$

1. $Incorrect$ $\text(Incorrect)$
2. $Correct$ $\text(Correct)$

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Trigonometric Identities

\sec θ = \frac{1}{cos θ}

$\sec\theta=\frac{1}{\text{cos}\theta}$

\csc θ = \frac{1}{sin θ}

$\csc\theta=\frac{1}{\text{sin}\theta}$

\cot θ = \frac{1}{tan θ}

$\cot\theta=\frac{1}{\text{tan}\theta}$

\tan θ = \frac{sin θ}{cos θ}

$\tan\theta=\frac{\text{sin}\theta}{\text{cos}\theta}$

\cot θ = \frac{cos θ}{sin θ}

$\cot\theta=\frac{\text{cos}\theta}{\text{sin}\theta}$

Work on changing the left side of the equation to make it equal to the value of the right side

	$\frac{{sin}^{2} θ - {tan}^{2} θ}{1 - {sec}^{2} θ}$ $(\text(sin)^2theta-\text(tan)^2theta)/(1-\text(sec)^2theta)$

$=$ $=$	$\frac{{sin}^{2} θ - {tan}^{2} θ}{- {tan}^{2} θ}$ $(\text(sin)^2theta-\text(tan)^2theta)/(-\text(tan)^2theta)$	$1 - {sec}^{2} θ = - {tan}^{2} θ$ $1-\text(sec)^2theta=-\text(tan)^2theta$

$=$ $=$	$\frac{{sin}^{2} θ}{- {tan}^{2} θ} - \frac{{tan}^{2} θ}{- {tan}^{2} θ}$ $(\text(sin)^2theta)/(-\text(tan)^2theta)-(\text(tan)^2theta)/(-\text(tan)^2theta)$	Separate the numerators

$=$ $=$	$\frac{{sin}^{2} θ}{\frac{{sin}^{2} θ}{{cos}^{2} θ}} + 1$ $(\text(sin)^2theta)/((\text(sin)^2theta)/(\text(cos)^2theta))+1$	${tan}^{2} θ = \frac{{sin}^{2} θ}{{cos}^{2} θ}$ $\text(tan)^2theta=(\text(sin)^2theta)/(\text(cos)^2theta)$

$=$ $=$	$- {cos}^{2} θ + 1$ $-\text(cos)^2theta+1$
$=$ $=$	$1 - {cos}^{2} θ$ $1-\text(cos)^2theta$
$=$ $=$	${sin}^{2} θ$ $\text(sin)^2theta$	$1 - {cos}^{2} θ = {sin}^{2} θ$ $1-\text(cos)^2theta=\text(sin)^2theta$

This means that

\frac{{sin}^{2} θ - {tan}^{2} θ}{1 - {sec}^{2} θ} = {sin}^{2} θ

$(\text(sin)^2theta-\text(tan)^2theta)/(1-\text(sec)^2theta)=\text(sin)^2theta$

Correct

$\text(Correct)$

Proving Trigonometric Identities

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1. Question

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Great Work!

Trigonometric Identities

2. Question

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Correct!

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3. Question

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Keep Going!

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4. Question

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Fantastic!

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5. Question

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Excellent!

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6. Question

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Nice Job!

Trigonometric Identities

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