If in a $\Delta ABC$, $\sqrt 3 \sin C = 2\sec A - \tan A$ and $\angle C = {\lambda ^0}$, find the value of $\lambda$

Question

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

Answer 1

Hint:

We have $$\sqrt3\sin C\cos A+\sin A=2$$

But $$\sqrt3\sin C\cos A+\sin A\le\sqrt{(\sqrt3\sin C)^2+1^2}=\sqrt{4-3\cos^2C}\le2$$

The equality occurs if $\cos C=0$

Answer 2

Use Weierstrass substitution

$$\sqrt3\sin C=\dfrac{2(1+t^2)-2t}{1-t^2}$$

$$\iff t^2(2+\sqrt3\sin C)-2t+2-\sqrt3\sin C=0$$

The discriminant $=2^2-4(2^2-3\sin^2C)=-12\cos^2C$ which is $\le0$

But we need $-12\cos^2C\ge0\implies12\cos^2C=0$

Answer 3

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 $\sin A=\frac {1}{2}.$

Answer 4

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}.$