If angles of a $\triangle{ABC}$ are in A.P. then find the value of : $$\dfrac{a}{c}\sin 2A+\dfrac{c}{a}\sin 2C$$ where $a,b,c$ are sides and $A,B,C$ are angles.
My attempts:
Let $\angle A =x-d , \angle B=x, \angle C=x+d\ \implies \angle B=60^{o}$.
So one possibility is following:
- $\angle A=30^{o}$
- $\angle C=90^{o}$
By sine law: $\dfrac{a}{\sin A}=\dfrac{b}{\sin B}=\dfrac{c}{\sin C}=2R$
$\implies \dfrac{a}{c}\sin 2A+\dfrac{c}{a}\sin 2C=\dfrac{\sin^{2}A \sin 2A+ \sin^{2} C \sin 2C}{\sin A \sin C} $ By value of $\angle A\ ,\angle C$.
Answer comes to be $\dfrac{\sqrt{3}}{4}$.
Is it correct, if not then where I go wrong? And I don't want to prove this by an assumption that $\angle A=30, \angle C=90$. How to do this without assuming angles.
Edit as suggested by @rohan :
original question is follows:
If angles of a $\triangle{ABC}$ are in A.P. then find the value of : $$\dfrac{a}{c}\sin 2C+\dfrac{c}{a}\sin 2A$$ where $a,b,c$ are sides and $A,B,C$ are angles.
I earnestly apologise for wasting your precious time, it was not deliberate.
I think there is a typo in your question. It should be to find the value of $\frac{a}{c}\sin 2C + \frac{c}{a}\sin 2A$.
Assuming the case it is so, then the problem simplifies to: $$\frac{\sin A}{\sin C}(2\sin C\cos C) + \frac{\sin C}{\sin A}(2\sin A\cos A) = 2\sin A\cos C + 2\sin C\cos A = 2\sin (A+C) = 2\sin 2B = 2\times \frac{\sqrt{3}}{2} =\sqrt{3}$$
The problem with the question as it is is the fact that it allows multiple possibilities.