Question

 

Answer 1

\(=\dfrac{2\cos^2\alpha-\sin^2\alpha-\cos^2\alpha}{\sin\alpha+\cos\alpha}+\sin\alpha\\ =\dfrac{\cos^2\alpha-\sin^2\alpha}{\sin\alpha+\cos\alpha}+\sin\alpha\\ =\dfrac{\left(\cos\alpha-\sin\alpha\right)\left(\cos\alpha+\sin\alpha\right)}{\cos\alpha+\sin\alpha}+\sin\alpha\\ =\cos\alpha-\sin\alpha+\sin\alpha=\cos\alpha\)