If in a $\Delta ABC$, $\sqrt 3 \sin C = 2\sec A - \tan A$ and $\angle C = {\lambda ^0}$, find the value of $\lambda$ If in a $\Delta ABC$, $\sqrt 3 \sin C = 2\sec A - \tan A$ and $\angle C = {\lambda ^0}$, find the value of $\lambda$
My approach is as follow
$\sqrt 3 \sin C = 2\sec A - \tan A \Rightarrow \sqrt 3 \sin C = \sec A + \sec A - \tan A$
$\sqrt 3 \sin C = \sec A + \left( {\frac{{{{\sec }^2}A - {{\tan }^2}A}}{{\sec A + \tan A}}} \right) \Rightarrow \sqrt 3 \sin C = \sec A + \left( {\frac{1}{{\sec A + \tan A}}} \right)$
$\sqrt 3 \sin C = \sec A + \left( {\frac{{\cos A}}{{1 + \sin A}}} \right) \Rightarrow \sqrt 3 \sin C = \frac{1}{{\cos A}} + \left( {\frac{{\cos A}}{{1 + \sin A}}} \right) \Rightarrow \sqrt 3 \sin C = \frac{{1 + \sin A + 1 - {{\sin }^2}A}}{{\cos A\left( {1 + \sin A} \right)}}$
$\sqrt 3 \sin C = \frac{{2 + \sin A - {{\sin }^2}A}}{{\cos A\left( {1 + \sin A} \right)}}$
Not able to proceed further

Another approach:
Assume $0\le x\lt1$, then:
$$4x^2+1\ge4x\implies 4-4x+x^2\ge 3-3x^2\implies \frac {(2-x)^2}{1-x^2}\ge3\implies \frac {2-x}{\sqrt {1-x^2}}\ge \sqrt3;$$
notice the equality holds when $x=\frac{1}{2}.$ Now, we have:
$$\sqrt3 \sin C=\frac{2}{cos A}-\frac{\sin A}{\cos A}=\frac {2-\sin A}{\sqrt {1-\sin^2 A}}.$$
Hence, by putting $x=\sin A$, we come to the conclusion that $\sin C=1$ and $\sin A=\frac {1}{2}.$
