中學數學題。超急! 謝謝謝謝！

Alternatively

1/sinθ – cosθ/ tanθ

= 1/sinθ – cosθ/(sinθ/cosθ) (Note: tanθ = sinθ/cosθ)

= 1/sinθ – cosθ cosθ/sinθ)

1 – cosθcosθ
= ----------------------- (Note: sin^2(θ) + cos^2(θ) = 1)
sin θ

sinθ sinθ
= ---------------
sin θ

= sinθ

= 3/5 (sinθ = 3/5 given)

1/sinθ – cosθ/ tanθ = 3/5 (Answer)

Comment:
You don’t have to calculate what cos θ and tan θ are)

sinθ=對邊/斜邊
cosθ=鄰邊/斜邊
tanθ=對邊/鄰邊

5^2-3^2=4^2

sinθ=3/5
cosθ=4/5
tanθ=3/4

1/sinθ- cosθ/tanθ
=1/(3/5)-(4/5)/(3/4)
=5/3-16/15
=3/5