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secondary 4 | A Maths
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idununderstandasinglething
Idununderstandasinglething

secondary 4 chevron_right A Maths chevron_right Singapore

please help with 2B and C thank you

Date Posted: 1 week ago
Views: 10
J
J
1 week ago
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)
J
J
1 week ago
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

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