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Trigonometric Identity ProofsTrigonometric 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`Hint
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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`Hint
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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`Hint
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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 sideFirst 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`Hint
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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 sideFirst 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)`Hint
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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`Hint
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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)`