Question

Evaluate, without using trigonometric tables:
$\frac{{\sin }^2\theta +{\sin }^2({90}^o-\theta )}{3({\sec }^2{61}^o-{\cot }^2{29}^o)}-\frac{3{\cot }^2{30}^o.{\sin }^2{54}^o.{\sec }^2{36}^o}{2({\csc }^2{65}^o-{\tan }^2{25}^{}o)}$

Answer 1

$\frac{{\sin }^2\theta +{\sin }^2({90}^o-\theta )}{3({\sec }^2{61}^o-{\cot }^2{29}^o)}-\frac{3{\cot }^2{30}^o.{\sin }^2{54}^o.{\sec }^2{36}^o}{2({\csc }^2{65}^o-{\tan }^2{25}^o)}$

$=\frac{{\sin }^2\theta +{\cos }^2\theta }{3({\sec }^2{61}^o-{\tan }^2{61}^o)}-\frac{3\times 3.{\sin }^2{54}^o.{\csc }^2{54}^o}{2({\csc }^2{65}^o-{\cot }^2{65}^o)}$

$=\frac{1}{3}-\frac{9}{2}$

$=\frac{2-27}{6}=-\frac{25}{6}$