MATHS]]]要STEP

11.
tan15tan75 +sin15/cos75
=tan15tan(90 - 15) + sin15/sin(90 - 75)
=1 + sin15/sin15
=2

12.
sin(90-θ)/tan(90-θ)
=cosθ / (1 / tanθ)
=cosθtanθ
=cosθ(sinθ/cosθ)
=sinθ

13.cos(90 - θ) x cosθ/tanθ
=sinθ x cosθ/ (sinθ/cosθ)
=sinθ x cos^2 θ/sinθ
=cos^2 θ

14.1-cos^2 θ/tan^2 (90'-θ)
=sin^2 θ/(1/tan^2 θ)
=sin^2 θ x tan^2 θ
= sin^2 θ x sin^2 θ/cos^2 θ
=sin^4 θ/cos^2 θ


那係恆等式? i 1/cosθ-cosθ=sinθtanθ ii 1-cosθ=cosθ/ta(90'-θ) iii 1/sin^2 θ-1/tan^2 θ=1
A.i B.i&ii C.i&iii D.ii&iii <--- Is this question 15?

i. LHS=1/cosθ - cosθ
=(1-cos^2 θ)/cosθ
=sin^2 θ/cosθ
=sinθ (sinθ/cosθ)
=sinθtanθ = RHS

ii. LHS =1 - cosθ
RHS = cosθ/tan(90 - θ)
=cosθtanθ
=sinθ
=/= LHS

iii.LHS=1/sin^2 θ - 1/tan^2 θ
=1/sin^2 θ - cos^2 θ/sin^2 θ
= (1- cos^2 θ)/sin^2 θ
=sin^2 θ/sin^2 θ
=1=RHS

Answer is C i&iii.

A,B,C係3個城市,已知 ...
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