Question

Without using trigonometric tables,find the value of the following:
${\left(\frac{tan{20}^{\circ }}{cosec{70}^{\circ }}\right)}^2+{\left(\frac{cot{20}^{\circ }}{sec{70}^{\circ }}\right)}^2+2tan{15}^{\circ }tan{45}^{\circ }tan{75}^{\circ }$

Answer 1

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

$={\left(\frac{\tan ({90}^{\circ }-{70}^{\circ })}{\csc {70}^{\circ }}\right)}^2+{\left(\frac{\cot ({90}^{\circ }-{70}^{\circ })}{\sec {70}^{\circ }}\right)}^2+2\tan ({90}^{\circ }-{75}^{\circ })\tan {45}^{\circ }\tan {75}^{\circ }$

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

$={\left(\cot {70}^{\circ }\sin {70}^{\circ }\right)}^2+{\left(\tan {70}^{\circ }\cos {70}^{\circ }\right)}^2+2\times 1\times 1$ since $\tan \theta \cot \theta =1$

$={(\frac{\cos {70}^{\circ }}{\sin {70}^{\circ }}\times \sin {70}^{\circ })}^2+{(\frac{\sin {70}^{\circ }}{\cos {70}^{\circ }}\times \cos {70}^{\circ })}^2+2$

$={\cos }^2{70}^{\circ }+{\sin }^2{70}^{\circ }+2$

$=1+2=3$