<s>[INST] 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
 [/INST] We had $\sqrt{3}\sin C\cos A+\sin A=2$.
It is well-known that the maximum of $a\cos\theta+b\sin\theta$ is $\sqrt{a^2+b^2}$ (and also its minumum is $-\sqrt{a^2+b^2}$).
So, the maximum of $\sqrt{3}\sin C\cos A+\sin A$ is $\sqrt{3\sin^2 C+1}$. It is clear that $\sqrt{3\sin^2 C+1}$ can be at most $2$ and it occurs at $C=90^{\circ}$ in the interval $(0^{\circ},180^{\circ})$. So, $\sqrt{3}\sin C\cos A+\sin A=2$ only if $C=90^{\circ}$.
We also can find $A$: $\sqrt{3}\cos A+\sin A=2\implies \sqrt{3}\cos A=2-\sin A\implies 3\cos^2A=4-4\sin A+\sin^2A\implies$$ 4\sin^2A-4\sin A+1=0\implies (2\sin^2 A-1)=0\implies \sin A=\frac{1}{2}\implies A=30^{\circ}.$
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