Question

${\cos }^2{0}^{\circ }-2{\cos }^2{30}^{\circ }+3{\cos }^2{90}^{\circ }=2({\sec }^2{45}^{\circ }-{\tan }^2{60}^{\circ })$

Answer 1

LHS $=$ ${\cos }^2{0}^0-2{\cos }^2{30}^0+3{\cos }^2{90}^0$

$=1-2.3/2+0=1-3=-2$

RHS $=2({\sec }^2{45}^0-{\tan }^2{60}^0)$

$=2(\frac{1}{2}-\frac{3}{2})=2(-1)=-2=$ LHS