Question

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

Answer 1

$\sin {39}^o=\sin (90-51{)}^o=\cos {51}^o$
$\sin {21}^o=\sin (90-69{)}^o=\cos {69}^o$
$\tan {11}^o=\tan (90-79{)}^o=\cot {79}^o$
$\tan {31}^o=\tan (90-59{)}^o=\cot {59}^o$
So, $\frac{\sin {39}^o}{\cos {51}^o}+2\tan {11}^o.\tan {31}^o.\tan {45}^o.\tan {59}^o.\tan {79}^o-2({\sin }^2{21}^o+{\sin }^2{69}^o)$
$=\frac{\cos {51}^o}{\cos {51}^o}+2\cot {79}^o.\cot {59}^o.\tan {45}^o.\tan {59}^o.\tan {79}^o-3({\cos }^{2}69+{\sin }^2{69}^o)$
$=1+2(\tan {79}^o.\cot {79}^o)(\tan {59}^o.\cot {59}^o).\tan {45}^o-3\left(1\right)$
$=1+2\left(1\right)\left(1\right)\left(1\right)-3$
$=3-3$
$=0$