Question

高中数学三角函数求值题，如下图，求详解，谢谢

Answer 1

首纳袭先根据已知条件求出角的正切值，然后化简所要森缺求的式子。洞春兄

Answer 2

两边同乘(1-tanθ)：tanθ=1-tanθ
2tanθ=1，则tanθ=1/2
(1)上下同除以cosθ：
原式=(sinθ/cosθ - 1)/(sinθ/cosθ + 1)
=(tanθ - 1)/(tanθ + 1)
=(1/2 - 1)/(1/2 + 1)
=(-1/2)/(3/2)=-1/3
(2)由诱导公式得：祥仿
原式=sinθ•(-cosθ) - (-sinθ)² - 2
=-sinθcosθ - sin²θ - 2
=-(1/2)•2sinθcosθ - (1-cos2θ)/2 - 2
=(-1/2)sin2θ + (1/2)cos2θ - 1/谨咐纤2 - 2
=(-1/2)•[2tanθ/(1+tan²θ)] + (1/2)•[(1-tan²简宴θ)/(1+tan²θ)] - 5/2
=(-1/2)•[1/(1 + 1/4)] + (1/2)•[(1 - 1/4)/(1 + 1/4)] - 5/2
=(-1/2)•(4/5) + (1/2)•(3/5) - 5/2
=-2/5 + 3/10 - 5/2
=-13/5