Prove that 1+sinθ−cosθ1+sinθ+cosθ=tan(θ2)
Also it is a question of S.L. Loney's Plane Trignonometry
What I've tried by now:
=1+sinθ−sin(90−θ)1+cos(90−θ)+cosθ=1+2cos45∘sin(θ−45∘)1+2cos45∘cos(45−θ)
Cause I do know
sinc+sind=2sin(c+d2)cos(c−d2)and cosc+cosd=2cos(c+d2)sin(c−d2)
I can't think of what to do next..
Answer
Let θ=2ϕ, then the thing to be proven is:
Prove that 1+sin(2ϕ)−cos(2ϕ)1+sin(2ϕ)+cos(2ϕ)=tan(ϕ)
Then use:
sin(2ϕ)=2sinϕcosϕ
cos(2ϕ)=cos2ϕ−sin2ϕ
and:
sin2ϕ+cos2ϕ=1
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