Question

Find the value of $4{\cos }^2{30}^o-3{\sec }^2{30}^o+2{\tan }^2{45}^o$.

Answer 1

Now,

$4{\cos }^2{30}^o-3{\sec }^2{30}^o+2{\tan }^2{45}^o$

$=4.\frac{3}{4}-3.\frac{4}{3}+2$ [ Using value of $\cos {30}^o,\sec {30}^o,\tan {45}^o$]

$=3-4+2$

$=1$.