Question

Prove with the help of trigonometric Identities.
$2{\sin }^2\theta +4{\sec }^2\theta +5{\cot }^2\theta +2{\cos }^2\theta -4{\tan }^2\theta -5{\csc }^2\theta =1$

Answer 1

$2{\sin }^2\theta +4se{c}^2\theta +5co{t}^2\theta +2co{s}^2\theta -4ta{n}^2\theta -5cose{c}^2\theta =2(si{n}^2\theta +cos62\theta )+4(se{c}^2\theta -ta{n}^2\theta )+5(co{t}^2\theta -cose{c}^2\theta )=2\times 1+4\times 1+5\times (-1)=2+4-5=1=RHS$