Question

Without using the trigonometrical tables, find the value of

$\frac{\sin {39}^{\circ }}{\cos {51}^{\circ }}+2\tan {11}^{\circ }\tan {31}^{\circ }\tan {45}^{\circ }\tan {59}^{\circ }\tan {79}^{\circ }-3\left({\sin }^2{21}^{\circ }+{\sin }^2{69}^{\circ }\right)$.

Answer 1

Solve the given expression.

Given that $\frac{\sin {39}^{\circ }}{\cos {51}^{\circ }}+2\tan {11}^{\circ }\tan {31}^{\circ }\tan {45}^{\circ }\tan {59}^{\circ }\tan {79}^{\circ }-3\left({\sin }^2{21}^{\circ }+{\sin }^2{69}^{\circ }\right)$

$$\begin{gathered} =\frac{\sin {39}^{\circ }}{\sin ({90}^{\circ }-{39}^{\circ })}+2\tan {11}^{\circ }\times \tan {31}^{\circ }\times 1\times \cot ({90}^{\circ }-{31}^{\circ })\times \cot ({90}^{\circ }-{11}^{\circ })-3({\sin }^2{21}^{\circ }+{\cos }^2{21}^{\circ }) \\ =\frac{\sin {39}^{\circ }}{\sin {39}^{\circ }}+2\tan {11}^{\circ }\times \tan {31}^{\circ }\times 1\times \cot {31}^{\circ }\times \cot {11}^{\circ }-3\times 1 \\ =1+2\tan {11}^{\circ }\times \tan {31}^{\circ }\times 1\times \frac{1}{\tan {31}^{\circ }}\times \frac{1}{\tan {11}^{\circ }}-3 \\ =1+2-3 \\ =0 \end{gathered}$$

Hence, the value of the $\frac{\sin {39}^{\circ }}{\cos {51}^{\circ }}+2\tan {11}^{\circ }\tan {31}^{\circ }\tan {45}^{\circ }\tan {59}^{\circ }\tan {79}^{\circ }-3\left({\sin }^2{21}^{\circ }+{\sin }^2{69}^{\circ }\right)$ is $0$.