a) tan²θ=(sinθ/cosθ)²=sin²θ/cos²θ
so tan²θ/(1+tan²θ)=(sin²θ/cos²θ)/(1+(sin²θ/cos²θ))=sin²θ/(cos²θ+sin²θ)=sin²θ/1=sin²θ QED.
b) sin²θ+2cosθ-1=2cosθ-(1-sin²θ)=2cosθ-cos²θ=cosθ(2-cosθ)≠cosθ. So there is a mistake in the question. It should have been:
Prove sin²θ+2cos²θ-1=cos²θ.
The given alleged identity is only true for specific values of θ, e.g., 0 and 90°.