To Prove:
tanθ + secθ = 1+sinθ/cosθ
and
cos²(90-θ)/1+cosθ = 1-cosθ
Please don't just give the answers, I actually don't know how to work it out.
Please explain.
Much appreciated.
tanθ + secθ = 1+sinθ/cosθ
and
cos²(90-θ)/1+cosθ = 1-cosθ
Please don't just give the answers, I actually don't know how to work it out.
Please explain.
Much appreciated.
-
tanθ=sinθ/cosθ and secθ =1/cosθ
now tanθ + secθ =sinθ/cosθ +1/cosθ
it implies sinθ+1/cosθ hence proved
and
cos²(90-θ)=sin²θ
so cos²(90-θ)/1+cosθ = sin²θ/1+cosθ
because sin²θ+cos²θ=1 it implies sin²θ=1-cos²θ
now 1-cos²θ/1+cosθ
(1-cosθ)(1+cosθ)/1+cosθ=1-cosθ hence proved
now tanθ + secθ =sinθ/cosθ +1/cosθ
it implies sinθ+1/cosθ hence proved
and
cos²(90-θ)=sin²θ
so cos²(90-θ)/1+cosθ = sin²θ/1+cosθ
because sin²θ+cos²θ=1 it implies sin²θ=1-cos²θ
now 1-cos²θ/1+cosθ
(1-cosθ)(1+cosθ)/1+cosθ=1-cosθ hence proved