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If in a $\Delta ABC$, $\angle B = \frac{2 \pi}{3}$, then $\cos A + \cos C$ lies in

1. $[\:- \sqrt{3}, \sqrt{3}\:]$
2. $(\: – \sqrt{3}, \sqrt{3}\:]$
3. $(\:\frac{3}{2}, \sqrt{3}\:)$
4. $(\:\frac{3}{2}, \sqrt{3}\:]$

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0
which is the right option?

+1 vote

Answer: $\mathbf A$

We have:

$\mathrm {\angle A + \angle B + \angle C } = \pi \;\text{(By angle sum property of$\Delta$)}$

$\Rightarrow \mathrm{\angle A + \dfrac{2\pi}{3} + \angle C}=\dfrac{\pi}{3}$

Now, We know that:
$\mathrm{\cos A + \cos C = 2\cos\frac{A+C}{2}.\cos\frac{A-C}{2}}$

$\Rightarrow \mathrm{\cos A + \cos C = 2\cos\frac{\pi}{6}.\cos\frac{A-C}{2}} = 2\times \frac{\sqrt3}{2}\cos\frac{A-C}{2} = \sqrt3\cos\frac{A-C}{2}$

$\Rightarrow \mathrm{\cos A + \cos C=\sqrt3\cos\frac{A-C}{2}}$

We know that:

$-1 \le\cos \theta \le1, \;\forall \theta$

$\mathrm{\Rightarrow -1\le \cos\big (\frac{A-C}{2} \big )\le1}$

$\mathrm{\Rightarrow -\sqrt3\le \sqrt 3\cos\big (\frac{A-C}{2} \big )\le \sqrt3}$

$\Rightarrow \text {Range} = [-\sqrt3, \;\sqrt3]$

$\therefore \mathbf A$ is the right answer.

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edited by
0
Superb solution. 👌
+1

But a mistake in the line

It will be $\displaystyle \therefore -1\le \cos \left(\frac{\mathrm{A}-\mathrm{C}}{2}\right) \le1$

And in this line $\sqrt{3} \cos \left(\frac{\mathrm{A}-\mathrm{C}}{2}\right)$ is missing.

0
Yes :D

Thanks.
+1
Corrected!

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