Ta có: tan α = \(\frac{sin\alpha}{cos\alpha}\)

=> M = (1+ cos α)(1 + \(\frac{sin\alpha}{cos\alpha}\))

=> M = 1 + \(\frac{sin\alpha}{cos\alpha}\) + cos α + sin α

Ta có sin² α + cos² α = 1

=> \(\left(\frac{1}{3}\right)^2\) + cos² α =1

=> \(\frac{1}{9}\)  + cos² α = 1

=> cos² α = 1 - \(\frac{1}{9}\) = \(\frac{8}{9}\)

=> cos α = \(\frac{\sqrt{8}}{3}\)

thay vào biểu thức M

M = 1 + \(\frac{\frac{1}{3}}{\frac{\sqrt{8}}{3}}\) + \(\frac{\sqrt{8}}{3}\) + \(\frac{1}{3}\)

M = .....