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secondary 4 | A Maths
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Date Posted: 3 years ago
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(sec A + tan A - 1) / (1 - sec A + tan A)
= (sec A + tan A - 1) / (1 + tan A - sec A) × (1 + tan A + sec A) / (1 + tan A + sec A)
(Rationalising the denominator. You see 1 + tan A as one term and sec A as the other term)
= (sec A + tan A - 1)(1 + tan A + sec A) / (1 + tan
A - sec A)(1 + tan A + sec A)
Notice the numerator can be rearranged into the from (x + y)(x - y) where x = tan A + sec A, y = 1. So we can rewrite it as x² - y²
So what we did actually rationalised the numerator as well.
= [ (tan A + sec A)² - 1²] / [(1 + tan A)² - sec² A)]
= (tan²A + 2 tan A sec A + sec² A - 1) / (1 + 2tan A + tan² A - sec² A)
= (2tan² A + 2 tan A sec A) / (2 tan A)
(Recall that tan² A + 1 = sec² A so we can simplify a lot of things here)
= tan A + sec A
(Cancel out common factor of 2 tan A)
= sin A / cos A + 1 / cos A
= (1 + sin A) / cos A
(Proved)
= (sec A + tan A - 1) / (1 + tan A - sec A) × (1 + tan A + sec A) / (1 + tan A + sec A)
(Rationalising the denominator. You see 1 + tan A as one term and sec A as the other term)
= (sec A + tan A - 1)(1 + tan A + sec A) / (1 + tan
A - sec A)(1 + tan A + sec A)
Notice the numerator can be rearranged into the from (x + y)(x - y) where x = tan A + sec A, y = 1. So we can rewrite it as x² - y²
So what we did actually rationalised the numerator as well.
= [ (tan A + sec A)² - 1²] / [(1 + tan A)² - sec² A)]
= (tan²A + 2 tan A sec A + sec² A - 1) / (1 + 2tan A + tan² A - sec² A)
= (2tan² A + 2 tan A sec A) / (2 tan A)
(Recall that tan² A + 1 = sec² A so we can simplify a lot of things here)
= tan A + sec A
(Cancel out common factor of 2 tan A)
= sin A / cos A + 1 / cos A
= (1 + sin A) / cos A
(Proved)
Alternatively,
(sec A + tan A - 1) / (1 - sec A + tan A)
= (1/cos A + sin A / cos A - 1) / (1 - 1/cos A + sin A / cos A)
= (1 + sin A - cos A) / (cos A - 1 + sin A)
(Multiply both numerator and denominator by cos A)
= (1 + sin A - cos A)(1 + sin A + cos A) / (cos A + sin A - 1)(cos A + sin A + 1)
(Rationalising via multiplying both numerator and denominator by 1 + cos A + sin A)
= [(1 + sin A)² - cos² A] / [(cos A + sin A)² - 1²]
= (1 + 2 sin A + sin² A - cos² A) / (cos² A + 2 cos A sin A + sin² A - 1)
= (2sin² A + 2 sin A) / (2 cos A sin A)
(Using the property sin² A + cos² A = 1 to simplify)
= (sin A + 1) / cos A
(Dividing by common factor 2 sin A)
(sec A + tan A - 1) / (1 - sec A + tan A)
= (1/cos A + sin A / cos A - 1) / (1 - 1/cos A + sin A / cos A)
= (1 + sin A - cos A) / (cos A - 1 + sin A)
(Multiply both numerator and denominator by cos A)
= (1 + sin A - cos A)(1 + sin A + cos A) / (cos A + sin A - 1)(cos A + sin A + 1)
(Rationalising via multiplying both numerator and denominator by 1 + cos A + sin A)
= [(1 + sin A)² - cos² A] / [(cos A + sin A)² - 1²]
= (1 + 2 sin A + sin² A - cos² A) / (cos² A + 2 cos A sin A + sin² A - 1)
= (2sin² A + 2 sin A) / (2 cos A sin A)
(Using the property sin² A + cos² A = 1 to simplify)
= (sin A + 1) / cos A
(Dividing by common factor 2 sin A)
See 1 Answer
See question's comments section for two ways to prove.
As for the second part,
(sin A + 1) / cos A = 3 cos A
sin A + 1 = 3 cos² A
sin A + 1 = 3 - 3 sin² A
3 sin² A + sin A - 2 = 0
(3 sin A - 2)(sin A + 1) = 0
3 sin A = 2 or sin A = -1
sin A = ⅔ or sin A = -1
For sin A = ⅔, Basic angle = sin-¹ (⅔) ≈ 0.7297 rad
Since sin A is positive, look at first and second quadrant
A ≈ 0.7297 rad or A ≈ π - 0.7297 rad
A = 0.730 rad (3s.f) or A ≈ 2.412 rad = 2.41 rad (3s.f)
For sin A = -1, A = 3π/2 (you should know that the sin x graph cuts y = -1 at x = 3π/2 and this repeats every 2π radians)
As for the second part,
(sin A + 1) / cos A = 3 cos A
sin A + 1 = 3 cos² A
sin A + 1 = 3 - 3 sin² A
3 sin² A + sin A - 2 = 0
(3 sin A - 2)(sin A + 1) = 0
3 sin A = 2 or sin A = -1
sin A = ⅔ or sin A = -1
For sin A = ⅔, Basic angle = sin-¹ (⅔) ≈ 0.7297 rad
Since sin A is positive, look at first and second quadrant
A ≈ 0.7297 rad or A ≈ π - 0.7297 rad
A = 0.730 rad (3s.f) or A ≈ 2.412 rad = 2.41 rad (3s.f)
For sin A = -1, A = 3π/2 (you should know that the sin x graph cuts y = -1 at x = 3π/2 and this repeats every 2π radians)
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