If $\sqrt{a+b}\tan\large\frac{\theta}{2}$$=\sqrt{a-b}\tan\large\frac{ \phi}{2}$ then $(b-a\sec \theta)(b+a\cos\phi)$ has maximum value $\lambda$ then $\lambda$ is

Question

$(a)\;0\qquad(b)\;1\qquad(c)\;2\qquad(d)\;None\;of\;these$