中四Trigonometry問題

Simplify the following.

sin^2 θ / 1- cos θ
= sin²θ/1-cosθ
= (1-cos²θ )/(1-cosθ)
= (1-cosθ )(1+osθ)/(1-cosθ)
= 1+cosθ

sin^2 θ tan^2 θ - tan^2 θ
= tan²θ (sin²θ -1)
= - tan²θ (1-sin²θ)
= - tan²θ (cos²θ)
= - (sin²θ/cos²θ)/cos²θ
= -sin²θ

tan ( 180° - x ) sin ( 90° + x )
= - tanx sin(90°- x)
= - tanx cosx
= - (sinx/cosx) cosx
= - sinx


Prove the following identities.

1 - tan^2 A / 1+ tan^2 A = 2 cos^2 A -1

左面方程式
= 1 - tan^2 A / 1+ tan^2 A
= (1- sin²A/cos²A) / (1 + sin²A/cos²A )
= [ (cos²A - sin²A)/cos²A ] / [ (cos²A + sin²A) / cos²A ]
= (cos²A - sin²A) / (cos²A + sin²A)
= cos²A - sin²A
= cos²A - (1 - cos²A)
= 2 cos²A -1
= 右面方程式

sin A / 1 + cos A = 1/ sin A - 1/ tan A

右面方程式
= 1/ sin A - 1/ tan A
= 1/ sin A - 1/ (sin A / cos A)
= 1/ sin A - cosA/ sinA
= (1-cosA) / sinA
= [ (1-cosA)(1+cosA) ] / [sinA(1+cosA)]
= 1-cos²A / [sinA(1+cosA)]
= sin²A / [sinA(1+cosA)]
= sinA / (1+ cos A)
= 左面方程式