Question

Without using trigonometric tables, evaluate that :
$\frac{\sec {39}^{\circ }}{\cos ec{51}^{\circ }}+\frac{2}{\sqrt{3}}\tan {17}^{\circ }\tan {38}^{\circ }\tan {60}^{\circ }\tan {52}^{\circ }\tan {73}^{\circ }-3({\sin }^{}{31}^{\circ }+{\sin }^2{59}^{\circ })$

Answer 1

We have

$\frac{sec{39}^{\circ }}{coses{51}^{\circ }}+\frac{2}{\sqrt{3}}tan{17}^{\circ }tan{38}^{\circ }tan{60}^{\circ }tan{52}^{\circ }tan{73}^{\circ }-3(si{n}^{}{31}^{\circ }+si{n}^2{59}^{\circ })$

$=\frac{sec{39}^{\circ }}{cosec({90}^{\circ }-{39}^{\circ })}+\frac{2}{\sqrt{3}}tan{17}^{\circ }tan{38}^{\circ }tan{60}^{\circ }tan({90}^{\circ }-{38}^{\circ })tan({90}^{\circ }-{17}^{\circ })-3(si{n}^2{31}^{\circ }+si{n}^2({90}^{\circ }-{31}^{\circ })$

$=\frac{sec{39}^{\circ }}{sec{39}^{\circ }}+\frac{2}{\sqrt{3}}tan{17}^{\circ }tan{38}^{\circ }tan{60}^{\circ }cot{38}^{\circ }cot{17}^{\circ }-3(si{n}^2{31}^{\circ }+co{s}^2{31}^{\circ })$

$=\frac{sec{39}^{\circ }}{sec{39}^{\circ }}+\frac{2}{\sqrt{3}}tan{60}^{\circ }-3(si{n}^2{31}^{\circ }+co{s}^2{31}^{\circ })$

$=1+\frac{2}{\sqrt{3}}\times \sqrt{3}-3\times 1=1+2-3=0$