Question

Without using trigonometric table evaluate the following:
$\frac{\sec {39}^{\circ }}{\text{cosec}{51}^{\circ }}+\frac{2}{\sqrt{3}}\tan {17}^{\circ }\tan {38}^{\circ }\tan {60}^{\circ }\tan {52}^{\circ }\tan {73}^{\circ }-3({\sin }^2{31}^{\circ }+{\sin }^2{59}^{\circ })$

• A. $0$
• B. $3$
• C. $\sqrt{2}$
• D. $5\sqrt{7}$
  

Answer 1

Answer: (A)

$0$

We know that
$\tan \alpha .\tan \beta =1$
When $\alpha +\beta ={90}^0$
Hence simplifying we get
$\frac{\sin {51}^0}{\cos {39}^0}+\frac{2}{\sqrt{3}}.\tan {60}^0-3({\sin }^2{31}^0+{\sin }^2{59}^0)$
$=\frac{\sin {51}^0}{\sin {51}^0}+\frac{2}{\sqrt{3}}.\tan {60}^0-3({\cos }^2{59}^0+{\sin }^2{59}^0)$
$=1+2-3$
$=0$