Answer 1

Consider LHS:
\[\cos 20^\circ + \cos 100^\circ + \cos 140^\circ\]
\[ = 2\cos \left( \frac{20^\circ + 100^\circ}{2} \right) \cos \left( \frac{20^\circ - 100^\circ}{2} \right) + \cos 140^\circ \left\{ \because \cos A + \cos B = 2\cos\left( \frac{A + B}{2} \right)\cos\left( \frac{A - B}{2} \right) \right\}\]
\[ = 2\cos 60^\circ \cos \left( - 40^\circ \right) + \cos 140^\circ\]
\[ = 2 \times \frac{1}{2}\cos 40^\circ + \cos 140^\circ\]
\[ = \cos 40^\circ+ \cos 140^\circ\]
\[ = 2\cos \left( \frac{40^\circ + 140^\circ}{2} \right) \cos \left( \frac{40^\circ - 140^\circ}{2} \right)\]
\[ = 2 \cos 90^\circ \cos 50^\circ\]
\[ = 0\]
Hence, LHS = RHS.