Ask Singapore Homework?
Upload a photo of a Singapore homework and someone will email you the solution for free.
Question
secondary 4 | A Maths
2 Answers Below
Anyone can contribute an answer, even non-tutors.
please help with 2B and C thank you
Date Posted: 4 years ago
Views: 233
2b)
(1 + cotx) / (1 + tanx)
= (1 + cosx/sinx) / (1 + sinx/cosx)
= ((sinx + cosx)/sinx)/((cosx + sinx)/cosx)
(Make common denominator for each sides of the fraction)
= (sinx + cosx)/sinx × cosx/(sinx + cosx)
(Invert the fraction. ÷ a/b = × b/a )
= cosx/sinx
= 1/tanx
(since tanx = sinx/cosx, inverting the fraction gives us the result)
(1 + cotx) / (1 + tanx)
= (1 + cosx/sinx) / (1 + sinx/cosx)
= ((sinx + cosx)/sinx)/((cosx + sinx)/cosx)
(Make common denominator for each sides of the fraction)
= (sinx + cosx)/sinx × cosx/(sinx + cosx)
(Invert the fraction. ÷ a/b = × b/a )
= cosx/sinx
= 1/tanx
(since tanx = sinx/cosx, inverting the fraction gives us the result)
2c)
1 + (cos²x/(sinx - 1))
= (sinx - 1 + cos²x)/(sinx - 1)
(Make common denominator)
= (sinx - cos²x - sin²x + cos²x)/(sinx - 1)
(cos²x + sin²x = 1, so -1 = -cos²x - sin²x)
= (-sin²x + sinx)/(sinx - 1)
= -sinx(sinx - 1)/(sinx - 1)
= -sinx
1 + (cos²x/(sinx - 1))
= (sinx - 1 + cos²x)/(sinx - 1)
(Make common denominator)
= (sinx - cos²x - sin²x + cos²x)/(sinx - 1)
(cos²x + sin²x = 1, so -1 = -cos²x - sin²x)
= (-sin²x + sinx)/(sinx - 1)
= -sinx(sinx - 1)/(sinx - 1)
= -sinx
See 2 Answers
360 Tutoring Program