Trigonometric Identity Proofs

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Question 1 of 6

1. Question

Identify if the equation is correct

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

`\text(Correct)`
`\text(Incorrect)`

Hint

Help Video

Correct

Great Work!

Incorrect

Trigonometric Identities

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

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

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

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

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

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

	`(\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+\text(sin) theta`

`=`	`(\text(cos) theta(1+\text(sin) theta))/(1-\text(sin)^2 theta)`	Difference of two squares

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

`\text(sin)^2theta+\text(cos)^2theta`	`=`	`1`
`\text(sin)^2theta+\text(cos)^2theta` `-\text(sin)^2theta`	`=`	`1` `-\text(sin)^2theta`	Subtract `\text(sin)^2theta` from both sides
`\text(cos)^2theta`	`=`	`1-\text(sin)^2theta`

	`(\text(cos) theta(1+\text(sin) theta))/(1-\text(sin)^2 theta)`

`=`	`(\text(cos) theta(1+\text(sin) theta))/(\text(cos)^2theta)`	`1-\text(sin)^2theta=\text(cos)^2theta`

`=`	`(1+\text(sin) theta)/(\text(cos) theta)`	`(\text(cos) theta)/(\text(cos)^2theta)=1/(\text(cos) theta)`

`=`	`1/(\text(cos) theta)+(\text(sin) theta)/(\text(cos)theta)`	Separate the values of the numerator

`=`	`\text(sec)theta+(\text(sin) theta)/(\text(cos)theta)`	`1/(\text(cos) theta)=\text(sec)theta`

`=`	`\text(sec)theta+\text(tan)theta`	`(\text(sin) theta)/(\text(cos)theta)=\text(tan)theta`

This means that `(\text(cos) theta)/(1-\text(sin) theta)=\text(sec) theta+\text(tan) theta`

`\text(Correct)`

Question 2 of 6

2. Question

Identify if the equation is correct

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

`\text(Correct)`
`\text(Incorrect)`

Hint

Help Video

Correct

Correct!

Incorrect

Trigonometric Identities

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

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

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

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

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

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

	`(\text(sin) A)/(1+\text(cos) A)+(1+\text(cos) A)/(\text(sin) A)`

`=`	`(\text(sin)^2A+(1+\text(cos) A)^2)/((1+\text(cos) A)\text(sin) A)`	Apply the rule of adding fractions

`=`	`(\text(sin)^2A+(1+\text(cos) A)(1+\text(cos) A))/((1+\text(cos) A)\text(sin) A)`	Expand

`=`	`(\text(sin)^2A+1+2\text(cos) A+\text(cos)^2A)/((1+\text(cos) A)\text(sin) A)`

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

	`(\text(sin)^2A+1+2\text(cos) A+\text(cos)^2A)/((1+\text(cos) A)\text(sin) A)`

`=`	`(1+1+2\text(cos) A)/((1+\text(cos) A)\text(sin) A)`	`\text(sin)^2A+\text(cos)^2A=1`

`=`	`(2+2\text(cos) A)/((1+\text(cos) A)\text(sin) A)`	Combine like terms

`=`	`(2(1+\text(cos) A))/((1+\text(cos) A)\text(sin) A)`	Factor out `2`

`=`	`2/(\text(sin) A)`	`(1+\text(cos) A)/(1+\text(cos) A)=1`

`=`	`2\text(csc) A`	`1/(\text(sin) A)=\text(csc) A`

This means that `(sin A)/(1+cos A)+(1+cos A)/(sin A)=2 csc A`

`\text(Correct)`

Question 3 of 6

3. Question

Identify if the equation is correct

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

`\text(Correct)`
`\text(Incorrect)`

Hint

Help Video

Correct

Keep Going!

Incorrect

Trigonometric Identities

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

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

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

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

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

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

First term

	`(1+\text(cot) theta)/(\text(csc) theta)`

`=`	`1/(\text(csc) theta)(1+\text(cot) theta)`	Factorise

`=`	`\text(sin) theta(1+\text(cot) theta)`	`1/(\text(csc) theta)=\text(sin) theta`

`=`	`\text(sin) theta(1+(\text(cos) theta)/(\text(sin) theta))`	`\text(cot) theta=(\text(cos) theta)/(\text(sin) theta)`

`=`	`\text(sin) theta+(\text(sin) theta*\text(cos) theta)/(\text(sin) theta)`	Expand

`=`	`\text(sin) theta+\text(cos) theta`	`(\text(sin) theta)/(\text(sin) theta)=1`

Second term

	`(\text(sec) theta)/(\text(tan) theta+\text(cot) theta)`

`=`	`(1/(\text(cos) theta))/(\text(tan) theta+\text(cot) theta)`	`\text(sec) theta=1/(\text(cos) theta)`

`=`	`(1/(\text(cos) theta))/((\text(sin) theta)/(\text(cos) theta)+\text(cot) theta)`	`\text(tan) theta=(\text(sin) theta)/(\text(cos) theta)`

`=`	`(1/(\text(cos) theta))/((\text(sin) theta)/(\text(cos) theta)+(\text(cos) theta)/(\text(sin) theta)`	`\text(cot) theta=(\text(cos) theta)/(\text(sin) theta)`

`=`	`(1/(\text(cos) theta))/((\text(sin)^2theta+\text(cos)^2theta)/(\text(cos) theta \text(sin) theta)`	Apply rule of adding fractions to the denominator

`=`	`(1/(\text(cos) theta))/(1/(\text(cos) theta \text(sin) theta)`	`\text(sin)^2theta+\text(cos)^2theta=1`

`=`	`1/(\text(cos) theta)*(\text(cos) theta \text(sin) theta)/1`	Apply rule of dividing fractions

`=`	`\text(sin) theta`

Finally, combine the two terms and subtract

		`\text(sin) theta+\text(cos) theta``-``\text(sin) theta`
	`=`	`\text(cos) theta`

This means that `(1+\text(cot) theta)/(\text(csc) theta)-(\text(sec) theta)/(\text(tan) theta+\text(cot) theta)=\text(cos) theta`

`\text(Correct)`

Question 4 of 6

4. Question

Identify if the equation is correct

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

`\text(Correct)`
`\text(Incorrect)`

Hint

Help Video

Correct

Fantastic!

Incorrect

Trigonometric Identities

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

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

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

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

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

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

First term

	`\text(tan) theta(1-\text(cot)^2theta)`

`=`	`\text(tan) theta-(\text(tan) theta*\text(cot)^2theta)`	Expand

`=`	`\text(tan) theta-((\text(sin) theta)/(\text(cos) theta)*\text(cot)^2theta)`	`\text(tan) theta=(\text(sin) theta)/(\text(cos) theta)`

`=`	`\text(tan) theta-((\text(sin) theta)/(\text(cos) theta)*(\text(cos)^2theta)/(\text(sin)^2theta))`	`\text(cot)^2theta=(\text(cos)^2theta)/(\text(sin)^2theta)`

`=`	`\text(tan) theta-((\text(cos) theta)/(\text(sin) theta))`	Simplify

`=`	`\text(tan) theta-\text(cot) theta`	`(\text(cos) theta)/(\text(sin) theta)=\text(cot) theta`

Second term

	`\text(cot) theta(1-\text(tan)^2theta)`

`=`	`\text(cot) theta-(\text(cot) theta*\text(tan)^2theta)`	Expand

`=`	`\text(cot) theta-((\text(cos) theta)/(\text(sin) theta)*\text(tan)^2theta)`	`\text(cot) theta=(\text(cos) theta)/(\text(sin) theta)`

`=`	`\text(cot) theta-((\text(cos) theta)/(\text(sin) theta)*(\text(sin)^2theta)/(\text(cos)^2theta))`	`\text(tan)^2theta=(\text(sin)^2theta)/(\text(cos)^2theta)`

`=`	`\text(cot) theta-((\text(sin) theta)/(\text(cos) theta))`	Simplify

`=`	`\text(cot) theta-\text(tan) theta`	`(\text(sin) theta)/(\text(cos) theta)=\text(tan) theta`

Finally, combine the two terms and add

		`\text(tan) theta-\text(cot) theta``+``\text(cot) theta-\text(tan) theta`
	`=`	`0`

This means that `\text(tan) theta(1-\text(cot)^2theta)+\text(cot) theta(1-\text(tan)^2theta)=0`

`\text(Correct)`

Question 5 of 6

5. Question

Identify if the equation is correct

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

`\text(Correct)`
`\text(Incorrect)`

Hint

Help Video

Correct

Excellent!

Incorrect

Trigonometric Identities

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

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

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

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

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

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

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

`=`	`(1)(2*1/(\text(sin)^2A)-1)`	`\text(csc)^2A=1/(\text(sin)^2A)`

`=`	`(2-\text(sin)^2A)/(\text(sin)^2A)`	Apply rule of subtracting fractions

`=`	`(2-(1-\text(cos)^2A))/(\text(sin)^2A)`	`\text(sin)^2A=1-\text(cos)^2A`

`=`	`(2-1+\text(cos)^2A)/(\text(sin)^2A)`	Adjust the subtracting values

`=`	`(1+\text(cos)^2A)/(\text(sin)^2A)`

This means that `\text(csc)^4A-\text(cot)^4A=(1+\text(cos)^2A)/(\text(sin)^2A)`

`\text(Correct)`

Question 6 of 6

6. Question

Identify if the equation is correct

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

`\text(Correct)`
`\text(Incorrect)`

Hint

Help Video

Correct

Nice Job!

Incorrect

Trigonometric Identities

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

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

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

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

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

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

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

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

`=`	`(\text(sin)^2theta)/(-\text(tan)^2theta)-(\text(tan)^2theta)/(-\text(tan)^2theta)`	Separate the numerators

`=`	`(\text(sin)^2theta)/((\text(sin)^2theta)/(\text(cos)^2theta))+1`	`\text(tan)^2theta=(\text(sin)^2theta)/(\text(cos)^2theta)`

`=`	`-\text(cos)^2theta+1`
`=`	`1-\text(cos)^2theta`
`=`	`\text(sin)^2theta`	`1-\text(cos)^2theta=\text(sin)^2theta`

This means that `(\text(sin)^2theta-\text(tan)^2theta)/(1-\text(sec)^2theta)=\text(sin)^2theta`

`\text(Correct)`

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