Question

Prove the following trigonometric identities:

$se{c}^6\theta =ta{n}^6\theta +3ta{n}^2\theta se{c}^2\theta +1$

Answer 1

We know that $se{c}^2\theta -ta{n}^2\theta =1$

Cubing both sides

=> $(se{c}^2\theta -ta{n}^2\theta {)}^3=1$

[Since $(a-b{)}^3=a^3-3a^{2}b+3a{b}^2-b^3$]

$se{c}^6\theta -3se{c}^4\theta ta{n}^2\theta +3se{c}^2\theta ta{n}^4\theta -ta{n}^6\theta =1$

$se{c}^6\theta =ta{n}^6\theta +3se{c}^4\theta ta{n}^2\theta -3se{c}^2\theta ta{n}^4\theta +1$

$se{c}^6\theta =ta{n}^6\theta +3se{c}^2\theta ta{n}^2\theta (se{c}^2\theta -ta{n}^2\theta )+1$

$se{c}^6\theta =ta{n}^6\theta +3se{c}^2\theta ta{n}^2\theta +1$

$Henceproved$