Question

Evaluate :$\frac{{\sec }^2({90}^o-\theta )-{\cot }^2\theta }{2(Si{n}^2{25}^o+{\sin }^2{65}^o)}+\frac{2{\sin }^2{30}^o.{\tan }^2{32}^o.{\tan }^2{58}^o}{3(Se{c}^2{33}^o-Co{t}^2{57}^o)}$

Answer 1

$\frac{{\sec }^2({90}^o-\theta )-{\cot }^2\theta }{2({\sin }^2{25}^o+{\sin }^2{65}^0)}+\frac{2{\sin }^2{30}^o.{\tan }^2{32}^o.{\tan }^2{58}^o}{3({\sec }^2{33}^o+{\cot }^2{57}^o)}=\frac{{cosec}^2\theta -{\cot }^2\theta }{2({\cos }^2{65}^o+{\sin }^2{65}^o)}+\frac{2{\left(\frac{1}{2}\right)}^2.{\cot }^2{58}^o.{\tan }^2{58}^o}{3({cosec}^2{57}^0-{\cot }^2{57}^o)}[\sec ({90}^o-x)=cosecx\sin ({90}^o-x)=\cos x\tan ({90}^o-x)=\cot x\sin {30}^o=\frac{1}{2}]$

$=\frac{1}{2}+\frac{1}{6}[1+{\cot }^{2}x={cosec}^{2}x\tan x=\frac{1}{\cot x}{\sin }^{2}x+{\cos }^{2}x=1]=\frac{2}{3}$