Prove that tan(π4+θ2)=1secθ−tanθ
Given: tan(π4+θ2)
We know, tan(A+B)=tanA+tanB1−tanAtanB
→tan(π4+θ2)=tanπ4+tanθ21−tanπ4+tanθ2=1+tanθ21−tanθ2
=cosθ2+sinθ2cosθ2−sinθ2
Multiply denominator to numerator by 2cosθ2
→2cos2θ2+2sinθ2cosθ22cos2θ2−2sinθ2cosθ2
→1+cosθ+sinθ1+cosθ−sinθ×1+cosθ+sinθ1+cosθ+sinθ
→(1+2cosθ2sinθ2)cosθ2cosθcosθ2
→1+sinθcosθ
→secθ+tanθ×secθ−tanθsecθ+tanθ
→1secθ−tanθ