Question

The value of ${\cos }^2{30}^0-{\sin }^2{30}^0=\cos {60}^o$
If above statement is true then enter $1$ and if false then write $0$.

Answer 1

${\cos }^2{30}^0-{\sin }^2{30}^0=\cos {60}^0$
L.H.S.: $={\cos }^2{30}^0-{\sin }^2{30}^0$

= ${\left(\frac{\sqrt{3}}{2}\right)}^2-{\left(\frac{1}{2}\right)}^2$

= $\frac{3}{4}-\frac{1}{4}$

= $\frac{2}{4}$

= $\frac{1}{2}$

= $\cos {60}^{\circ }$

hence, L.H.S. = R.H.S.