Question

Without using trigonometric tables, find the value of the following expressions.
$\frac{\sec ({90}^o-\theta )\csc \theta -\tan ({90}^o-\theta )\cot \theta +{\cos }^2{25}^o+{\cos }^2{65}^o}{3\tan {27}^o.\tan {63}^o}$

Answer 1

We have,

$\frac{\sec \left({90}^{\circ }-\theta \right)\cos ec\theta -\tan \left({90}^{\circ }-\theta \right)\cot \theta +{\cos }^2{25}^{\circ }+{\cos }^2{65}^{\circ }}{3\tan {27}^{\circ }\tan {65}^{\circ }}$

$=\frac{\cos e{c}^2\theta -{\cot }^2\theta +{\cos }^2{25}^{\circ }+{\sin }^2{25}^{\circ }}{3\tan {27}^{\circ }\cot {27}^{\circ }}=\frac{1+1}{3}=\frac{2}{3}$

Hence, this is the answer.