Question

Evalute: $\frac{{\sin }^2\theta +{\sin }^2\left(90-\theta \right)}{3({\sec }^2{61}^{\circ }-{\cot }^2{29}^{\circ })}-\frac{3{\cot }^2{30}^{\circ }{\sin }^2{54}^{\circ }{\sec }^2{36}^{\circ }}{2\left(\cos e{c}^2{65}^{\circ }-{\tan }^2{25}^{\circ }\right)}$

Answer 1

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

$\Rightarrow$ $\frac{{\sin }^2\theta +{\sin }^2(90-\theta )}{3[{\sec }^2{61}^o-{\cot }^2(90-61{)}^o]}-\frac{3{\cot }^2{30}^o{\sin }^2{54}^o{\sec }^2(90-54{)}^o}{2[cose{c}^2{65}^o-{\tan }^2(90-65{)}^o]}$

Using quadrant concepts we get,

$\Rightarrow$ $\frac{{\sin }^2\theta +{\cos }^2\theta }{3[{\sec }^2{61}^o-{\tan }^2{61}^o]}-\frac{3{\cot }^2{30}^o{\sin }^2{54}^{o}cose{c}^2{54}^o}{2[cosec{65}^o-{\cot }^2{65}^o}$

Using formulas,

${\sin }^2\theta +co{s}^2\theta =1$

${\sec }^2\theta -{\tan }^2\theta =1$

$cose{c}^2\theta -{\cot }^2\theta =1$

$cosec\theta =\frac{1}{{\sin }^2\theta }$

We get,

$\Rightarrow$ $\frac{1}{3\left(1\right)}-\frac{3{\cot }^2{30}^o\left(1\right)}{2\left(1\right)}$

$\Rightarrow$ $\frac{1}{3}-\frac{3\times (\sqrt{3}{)}^2}{2}$

$\Rightarrow$ $\frac{2-27}{3\left(2\right)}$

$\Rightarrow$ $\frac{-25}{6}$