Question

Prove that :$\frac{\cos {70}^o}{\sin {20}^o}+\frac{\cos {59}^o}{\sin {31}^o}-8{\sin }^2{30}^o=0$

Answer 1

LHS :

$\frac{\cos {70}^o}{\sin {20}^o}+\frac{\cos {59}^o}{\sin {31}^o}-8{\sin }^2{30}^o$

$=\frac{\cos ({90}^o-{20}^o)}{\sin {20}^o}+\frac{\cos ({90}^o-{31}^o)}{\sin {31}^o}-8{\sin }^2{30}^o$

$=1+1-8(\frac{1}{2}{)}^2$

$=1+1-2=0$ = RHS