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Question 1 of 6
Identify if the equation is correct
cosθ1-sinθ=secθ+tanθ
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Work on changing the left side of the equation to make it equal to the value of the right side
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cosθ1-sinθ |
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cosθ⋅(1+sinθ)1−sinθ⋅(1+sinθ) |
Multiply the numerator and denominator by 1+sinθ |
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cosθ(1+sinθ)1-sin2θ |
Difference of two squares |
Recall that sin2θ+cos2θ=1. Derive this formula to further simplify the expression
sin2θ+cos2θ |
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1 |
sin2θ+cos2θ -sin2θ |
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1 -sin2θ |
Subtract sin2θ from both sides |
cos2θ |
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1-sin2θ |
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cosθ(1+sinθ)1-sin2θ |
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cosθ(1+sinθ)cos2θ |
1-sin2θ=cos2θ |
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1+sinθcosθ |
cosθcos2θ=1cosθ |
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1cosθ+sinθcosθ |
Separate the values of the numerator |
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secθ+sinθcosθ |
1cosθ=secθ |
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secθ+tanθ |
sinθcosθ=tanθ |
This means that cosθ1-sinθ=secθ+tanθ
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Question 2 of 6
Identify if the equation is correct
sinA1+cosA+1+cosAsinA=2cscA
Incorrect
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Work on changing the left side of the equation to make it equal to the value of the right side
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sinA1+cosA+1+cosAsinA |
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sin2A+(1+cosA)2(1+cosA)sinA |
Apply the rule of adding fractions |
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sin2A+(1+cosA)(1+cosA)(1+cosA)sinA |
Expand |
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sin2A+1+2cosA+cos2A(1+cosA)sinA |
Recall that sin2θ+cos2θ=1. Derive this formula to further simplify the expression
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sin2A+1+2cosA+cos2A(1+cosA)sinA |
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1+1+2cosA(1+cosA)sinA |
sin2A+cos2A=1 |
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2+2cosA(1+cosA)sinA |
Combine like terms |
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2(1+cosA)(1+cosA)sinA |
Factor out 2 |
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2sinA |
1+cosA1+cosA=1 |
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2cscA |
1sinA=cscA |
This means that sinA1+cosA+1+cosAsinA=2cscA
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Question 3 of 6
Identify if the equation is correct
1+cotθcscθ-secθtanθ+cotθ=cosθ
Incorrect
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Work on changing the left side of the equation to make it equal to the value of the right side
First term
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1+cot θcsc θ |
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1csc θ(1+cot θ) |
Factorise |
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sin θ(1+cot θ) |
1csc θ=sin θ |
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sin θ(1+cos θsin θ) |
cot θ=cos θsin θ |
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= |
sin θ+sin θ⋅cos θsin θ |
Expand |
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sin θ+cos θ |
sin θsin θ=1 |
Second term
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sec θtan θ+cot θ |
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1cos θtan θ+cot θ |
sec θ=1cos θ |
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1cos θsin θcos θ+cot θ |
tan θ=sin θcos θ |
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1cos θsin θcos θ+cos θsin θ |
cot θ=cos θsin θ |
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1cos θsin2θ+cos2θcos θ sin θ |
Apply rule of adding fractions to the denominator |
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1cos θ1cos θ sin θ |
sin2θ+cos2θ=1 |
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1cos θ⋅cos θ sin θ1 |
Apply rule of dividing fractions |
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sin θ |
Finally, combine the two terms and subtract
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sin θ+cos θ-sin θ |
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cos θ |
This means that 1+cot θcsc θ-sec θtan θ+cot θ=cos θ
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Question 4 of 6
Identify if the equation is correct
tanθ(1-cot2θ)+cotθ(1-tan2θ)=0
Incorrect
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Work on changing the left side of the equation to make it equal to the value of the right side
First term
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tan θ(1-cot2θ) |
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tan θ-(tan θ⋅cot2θ) |
Expand |
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tan θ-(sin θcos θ⋅cot2θ) |
tan θ=sin θcos θ |
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tan θ-(sin θcos θ⋅cos2θsin2θ) |
cot2θ=cos2θsin2θ |
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tan θ-(cos θsin θ) |
Simplify |
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tan θ-cot θ |
cos θsin θ=cot θ |
Second term
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cot θ(1-tan2θ) |
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cot θ-(cot θ⋅tan2θ) |
Expand |
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cot θ-(cos θsin θ⋅tan2θ) |
cot θ=cos θsin θ |
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cot θ-(cos θsin θ⋅sin2θcos2θ) |
tan2θ=sin2θcos2θ |
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cot θ-(sin θcos θ) |
Simplify |
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cot θ-tan θ |
sin θcos θ=tan θ |
Finally, combine the two terms and add
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tan θ-cot θ+cot θ-tan θ |
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0 |
This means that tan θ(1-cot2θ)+cot θ(1-tan2θ)=0
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Question 5 of 6
Identify if the equation is correct
csc4A-cot4A=1+cos2Asin2A
Incorrect
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Work on changing the left side of the equation to make it equal to the value of the right side
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csc4A-cot4A |
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(csc2A-cot2A)(csc2A+cot2A) |
Factorise |
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[csc2A-(csc2A-1)][csc2A+(csc2A-1)] |
cot2A=csc2A-1 |
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(csc2A-csc2A+1)(csc2A+csc2A-1) |
Adjust the subtracting values |
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(1)(2csc2A-1) |
Combine like terms |
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(1)(2⋅1sin2A-1) |
csc2A=1sin2A |
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2-sin2Asin2A |
Apply rule of subtracting fractions |
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2-(1-cos2A)sin2A |
sin2A=1-cos2A |
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2-1+cos2Asin2A |
Adjust the subtracting values |
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1+cos2Asin2A |
This means that csc4A-cot4A=1+cos2Asin2A
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Question 6 of 6
Identify if the equation is correct
sin2θ-tan2θ1-sec2θ=sin2θ
Incorrect
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Work on changing the left side of the equation to make it equal to the value of the right side
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sin2θ-tan2θ1-sec2θ |
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sin2θ-tan2θ-tan2θ |
1-sec2θ=-tan2θ |
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sin2θ-tan2θ-tan2θ-tan2θ |
Separate the numerators |
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sin2θsin2θcos2θ+1 |
tan2θ=sin2θcos2θ |
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-cos2θ+1 |
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1-cos2θ |
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sin2θ |
1-cos2θ=sin2θ |
This means that sin2θ-tan2θ1-sec2θ=sin2θ