Question

Prove the following:
$2se{c}^2\theta se{c}^4\theta -2cose{c}^2\theta +cose{c}^4\theta =co{t}^4\theta -ta{n}^4\theta$

Answer 1

Taking LHS
$2se{c}^2\theta se{c}^4\theta -2cose{c}^2\theta +cose{c}^4\theta se{c}^2\theta (2-se{c}^2\theta )-cose{c}^2\theta (2-cose{c}^2\theta )se{c}^2\theta \{2-(1+ta{n}^2\theta \left)\right\}-cose{c}^2\theta \{2-(1+co{t}^2\theta \left)\right\}se{c}^2\theta \{1-ta{n}^2\theta \}-cose{c}^2\theta \{1-co{t}^2\theta \}(1+ta{n}^2\theta )(1-tan62\theta )-(1+co{t}^2\theta )(1-co{t}^2\theta )1-ta{n}^4\theta -\{1-co{t}^4\theta \}=co{t}^4-ta{n}^4\theta$ = RHS