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Simplifying Trigonometric Identities>
Simplifying Trigonometric IdentitiesSimplifying Trigonometric Identities
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Question 1 of 7
1. Question
Which of the following are trigonometric identities?- 1.
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Trigonometric Identities
secθ=1cosθcscθ=1sinθcotθ=1tanθtanθ=sinθcosθcotθ=cosθsinθUsing the given image, we can see that the radius of the circle will be 1. Therefore, the circle follows the formula x2+y2=1, where 1 is the radius.Make a reference triangle by making a line from the point of origin (0,0) going to any point in the circle to represent the radius and connect it to either the x or y-axisNotice that, given angle θ, the circle has the vertical line parallel to the y-axis as its opposite side and the line on the x-axis as the adjacent sideNow that we have the opposite and adjacent sides and the radius 1 as the hypotenuse, we can use the trigonometric functions to find their respective valuessin = oppositehypotenuse = y1 = y cos = adjacenthypotenuse = x1 = x x and y are both coordinate values, so we can write the coordinates as (cosθ,sinθ)Finally, use the values of x and y and derive its trigonometric identitiesx2+y2 = 1 cos2θ+sin2θ = 1 Substitute known values (cos2θ+sin2θ)÷cos2θ = 1÷cos2θ Divide both sides by cos2θ 1+sin2θcos2θ = 1cos2θ 1+tan2θ = 1cos2θ sin2θcos2θ=tan2θ 1+tan2θ = sec2θ 1cos2θ=sec2θ sec2θ = 1+tan2θ x2+y2 = 1 cos2θ+sin2θ = 1 Substitute known values (cos2θ+sin2θ)÷sin2θ = 1÷sin2θ Divide both sides by sin2θ cos2θsin2θ+1 = 1sin2θ cot2θ+1 = 1sin2θ cos2θsin2θ=cot2θ cot2θ+1 = csc2θ 1sin2θ=csc2θ csc2θ = cot2θ+1 sec2θ=1+tan2θcsc2θ=cot2θ+1 -
Question 2 of 7
2. Question
Simplify11+sinθ+11-sinθ-
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Trigonometric Identities
secθ=1cosθcscθ=1sinθcotθ=1tanθtanθ=sinθcosθcotθ=cosθsinθApply the rule of adding fractions and simplify the expression11+sinθ+11-sinθ = 1-sinθ+1+sinθ(1+sinθ)(1-sinθ) = 2(1+sinθ)(1-sinθ) Combine like terms = 21-sin2θ Expand Recall that sin2θ+cos2θ=1. Derive this formula to further simplify the expressionsin2θ+cos2θ = 1 sin2θ+cos2θ -sin2θ = 1 -sin2θ Subtract sin2θ from both sides cos2θ = 1-sin2θ 21-sin2θ = 2cos2θ 1-sin2θ=cos2θ = 21×1cos2θ Take out 2 = 2×sec2θ 1cos2θ=sec2θ = 2sec2θ 2sec2θ -
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Question 3 of 7
3. Question
Simplifytanθsecθ-
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Chapters- Chapters
Trigonometric Identities
secθ=1cosθcscθ=1sinθcotθ=1tanθtanθ=sinθcosθcotθ=cosθsinθSubstitute known values to tan and sectan = sincos sec = 1cos tan θsec θ = sincos1cos Substitute known values = sin θ Apply rule of dividing fractions sin θ -
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Question 4 of 7
4. Question
Simplifysecθ-secθsin2θ-
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Trigonometric Identities
secθ=1cosθcscθ=1sinθcotθ=1tanθtanθ=sinθcosθcotθ=cosθsinθFactor out sec θ on both termssec θ-sec θ sin2θ = sec θ(1-sin2θ) Factorise Recall that sin2θ+cos2θ=1. Derive this formula to further simplify the expressionsin2θ+cos2θ = 1 sin2θ+cos2θ -sin2θ = 1 -sin2θ Subtract sin2θ from both sides cos2θ = 1-sin2θ sec θ(1-sin2θ) = sec θ(cos2θ) 1-sin2θ=cos2θ = 1cos θ⋅(cos2θ) sec θ=1cos θ = cos2θcos θ = cos θ cos θ -
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Question 5 of 7
5. Question
Simplifysec2A-csc2A-
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The difference of two squares is a format of a quadratic expression where there are two terms that are both perfect squares.First, find the square of the two terms.First termsec2A = sec A⋅sec A Second termcsc2A = csc A⋅csc A Finally, get the sum and difference of the two values.(sec A-csc A)(sec A+csc A) (sec A-csc A)(sec A+csc A) -
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Question 6 of 7
6. Question
Simplifytan2θsecθ-1-
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Need TextPlayCurrent Time 0:00/Duration Time 0:00Remaining Time -0:00Stream TypeLIVELoaded: 0%Progress: 0%0:00Fullscreen00:00MutePlayback Rate1x- 2x
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Chapters- Chapters
Trigonometric Identities
secθ=1cosθcscθ=1sinθcotθ=1tanθtanθ=sinθcosθcotθ=cosθsinθRecall that 1+tan2θ=sec2θ. Derive this formula to further simplify the expression1+tan2θ = sec2θ 1+tan2θ -1 = sec2θ -1 Subtract 1 from both sides tan2θ = sec2θ-1 tan2θsecθ-1 = sec2θ-1secθ-1 tan2θ=sec2θ-1 = (secθ-1)(secθ+1)secθ-1 Factorise = secθ+1 secθ-1secθ-1=1 secθ+1 -
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Question 7 of 7
7. Question
Simplify(tan2β)(cot2β+1)tan2β+1- (1)
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Chapters- Chapters
Trigonometric Identities
secθ=1cosθcscθ=1sinθcotθ=1tanθtanθ=sinθcosθcotθ=cosθsinθRecall that 1+cot2θ=csc2θ. Derive this formula to further simplify the expression(tan2β)(cot2β+1)tan2β+1 = tan2β⋅csc2θtan2β+1 1+cot2θ=csc2θ Finally, substitute known values to tan, csc, and sectan = sincos csc = 1sin sec = 1cos tan2β⋅csc2θtan2β+1 = sin2βcos2β⋅1csc2θ1cos2θ Substitute known values = sin2βcos2β⋅1csc2θ⋅cos2θ1 Apply rule of dividing fractions = 1 1