Question

Without using tables, prove the following $\sin {40}^o-\cos {70}^o=\sqrt{3}\cdot \cos {80}^o$.

Answer 1

L.H.S

$=\sin {40}^0-\cos {70}^0$

We know that,

$\sin ({90}^0-\theta )=\cos \theta$

Therefore,

$=\cos {50}^0-\cos {70}^0$

We know that,

$\cos C-\cos D=2\sin \frac{C+D}{2}.\sin \frac{D-C}{2}$

Therefore,

$=2\sin \frac{{50}^0+{70}^0}{2}.\sin \frac{{70}^0-{50}^0}{2}$

$=2\sin {60}^0.\sin {10}^0$

$=\sqrt{3}.\cos {80}^0$

Hence, proved.