Question

Without using trigonometrical tables, evaluate :
${\left(\frac{\tan {20}^0}{\csc {70}^0}\right)}^2+{\left(\frac{\cot {20}^0}{\sec {70}^{.}}\right)}^2+2\tan {15}^0\tan {45}^0\tan {75}^0$

Answer 1

Formula:

$\tan (90-x)=\cot x$

$\sec (90-x)=\csc x$

$\csc (90-x)=\sec x$

${\left(\frac{\tan {20}^{\circ }}{\csc {70}^{\circ }}\right)}^2+{\left(\frac{\cot {20}^{\circ }}{\sec {70}^{\circ }}\right)}^2+2\tan {15}^{\circ }\tan {45}^{\circ }\tan {75}^{\circ }$

$={\left(\frac{\tan {20}^{\circ }}{\sec {20}^{\circ }}\right)}^2+{\left(\frac{\cot {20}^{\circ }}{\csc {20}^{\circ }}\right)}^2+2\tan {15}^{\circ }\tan {45}^{\circ }\cot {15}^{\circ }$

$={\sin }^2{20}^{\circ }+{\cos }^2{20}^{\circ }+2\tan {15}^{\circ }\cot {15}^{\circ }\tan {45}^{\circ }$

$=1+2\left(1\right)\left(1\right)$

$=1+2$

$=3$(Ans)