Correct Answer - Option 2 :
\(\frac{3}{16}\)
Concept:
2cos A cos B = cos (A + B) + cos (A - B)
cos (180° - θ ) = - cos θ
Calculation:
cos 10° cos 30° cos 50° cos 70° .
= \(\rm \frac{1}{2}\ \cos 30^{0}\ \cos 50^{0}\ (2\cos 10^{0}\ \cos70^{0})\)
= \(\rm \frac{1}{2}\times\frac{\sqrt{3}}{2}\ \cos 50^{0}\left \{ \cos\ (70^{0}+ 10^{0})+ \cos\ (70^{0}- 10^{0}) \right \}\)
= \(\rm \frac{\sqrt{3}}{4}\ cos\ 50^{0}\left [ \cos80^{0}+ \frac{1}{2} \right ]\)
= \(\rm \frac{\sqrt{3}}{8}[ (2cos\ 50^{0}\ \cos80^{0}\ ) \ + \cos 50^{0}]\)
= \(\rm \frac{\sqrt{3}}{8}\ \left \{ \cos\ (80^{0}+ 50^{0})+ \cos\ (80^{0}- 50^{0}) + \cos\ 50^{0}\right \}\)
= \(\rm \frac{\sqrt{3}}{8}[\cos 130^{0} + \cos 30^{0} + \cos50^{0}]\)
= \(\rm \frac{\sqrt{3}}{8} [\cos \left ( 180^{0}- 50^{0} \right ) + \frac{\sqrt 3}{2} + \cos 50^{0}]\)
= \(\rm \frac{\sqrt{3}}{8} [ -\cos 50^{0}+ \frac{\sqrt 3}{2} + \cos 50^{0}]\)
= \(\frac{3}{16}\) .
The correct option is 2.
