Question

$\frac{5{\cos }^2{60}^o+4{\sec }^2{30}^o-\tan {45}^o}{{\sin }^2{30}^o+{\cos }^2{30}^o}$.

Answer 1

Note that, cos(60)= 1/2

cos(30)= √3/2

sin(30)= 1/2

tan(45)= 1

Now as per question

5cos²(60) +4sec²(30) - tan(45)

= 5$\ast {\frac{1}{2}}^2+4{\frac{2}{\surd 3}}^2-1$

=$\frac{5}{4}+\frac{16}{3}-1$

= $\frac{67}{12}$

And cos²x + sin²x = 1

Hence, Ans = $\frac{67}{12}$