Evaluate (5cos^2 60 + 4 sec^2 30 - tan ^2 45)/(sin^2 30 + cos^2 30

Q) Evaluate $\frac{5cos^2 60 + 4 sec^2 30 - tan ^2 45}{sin^2 30 + cos^2 30}$

Ans:

Since sin²θ + cos²θ = 1

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

= $\frac{5 cos^2 60 + 4 sec^2 30 - tan ^2 45}{1}$

= 5 cos² 60 + 4 sec² 30 – tan² 45

= $5 (\frac{1}{2})^2 + 4(\frac{2}{\sqrt3})^2 - (1)^2$

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

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

= $\frac{67}{12}$

Contact us