Question

Prove: $\sin {40}^o-\cos {70}^o=\sqrt{3}\cdot \cos {80}^o$.

Answer 1

Now,

$\sin {40}^o-\cos {70}^o$

$=\cos {50}^o-\cos {70}^o$ [ Since $\sin {40}^o=\cos {50}^o$]

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

$=2.\frac{\sqrt{3}}{2}.\cos {80}^o$. [ Since $\sin {10}^o=\cos {80}^o$]

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