Question

Without using trigonometric tables, evaluate the following :
$\sec {41}^o.\sin {49}^o+\cos {49}^o\csc {41}^o-\frac{2}{\sqrt{3}}(\tan {20}^o.\tan {60}^o.\tan {70}^o)$

Answer 1

$\sec {41}^{\circ }\sin {49}^{\circ }+\cos {49}^{\circ }\csc {41}^{\circ }-\frac{2}{\sqrt{3}}\left(\tan {20}^{\circ }\tan {60}^{\circ }\tan {70}^{\circ }\right)$

$=\sec {41}^{\circ }\sin (90-41)+\cos {41}^{\circ }\csc (90-49)-\frac{2}{\sqrt{3}}\left(\tan (90-70)\sqrt{3}\tan {70}^{\circ }\right)$

$=\sec {41}^{\circ }\cos {41}^{\circ }+\cos {41}^{\circ }\sec {41}^{\circ }-\frac{2}{\sqrt{3}}\left(\cot {70}^{\circ }\sqrt{3}\tan {70}^{\circ }\right)$

$=1+1-2=0$ using reciprocal relations of trigonometry.