Correct Answer - Option 4 : cot 6x
Concept:
- sin A + sin B = \(\rm 2sin\left ( \frac{A+B}{2} \right )\cos \left ( \frac{A-B}{2} \right )\)
- cos A + cos B = \(\rm 2cos\left ( \frac{A+B}{2} \right )\cos \left ( \frac{A-B}{2} \right )\)
Calculation:
We have , \(\rm \frac{\cos 7x\ +\ \cos 6x\ +\ \cos5x }{\sin 7x\ +\ \sin 6x\ +\ \sin 5x}\)
We know that,
sin A + sin B = \(\rm 2sin\left ( \frac{A+B}{2} \right )\cos \left ( \frac{A-B}{2} \right )\)
cos A + cos B = \(\rm 2cos\left ( \frac{A+B}{2} \right )\cos \left ( \frac{A-B}{2} \right )\)
using the above formula , we get
= \(\rm \frac{(\cos 7x\ +\ \cos 5x\ ) +\ \cos6x }{(\sin 7x\ +\ \sin 5x\ )+\ \sin 6x}\)
= \(\rm \frac{ 2cos\left ( \frac{7x+5x}{2} \right )\cos \left ( \frac{7x-5x}{2} \right )+\cos 6x}{2sin\left ( \frac{7x+5x}{2} \right )\cos \left ( \frac{7x-5x}{2} \right )+\sin 6x}\)
= \(\rm \frac{2 \cos 6x\ \cos x +\cos 6x }{2 \sin 6x\ \cos x +\sin6x}\)
= \(\rm \frac{\cos6x\ (2 \ cosx + 1)}{\sin6x\ (2 \ cosx + 1)}\)
= cot 6x .
The correct option is 4
