Question

Please prove that

Answer 1

Recall the Pythagorean identity.

$\sin^2(x) + \cos^2(x) = 1 \implies \left\langle \begin{matrix} \tan^2(x) + 1 = \sec^2(x) \\ 1 + \cot^2(x) = \csc^2(x) \end{matrix} \right.$

$\implies \sec^2(\beta) - \csc^2(\beta) = (\tan^2(\beta) + 1) - (1 + \cot^2(\beta)) \\\\ ~~~~~~~~= \tan^2(\beta) - \cot^2(\beta)$

Recall the difference of squares identity.

$a^2 - b^2 = (a - b) (a + b)$

$\implies \sec^2(\beta) - \csc^2(\beta) = (\tan(\beta) - \cot(\beta)) (\tan(\beta) + \cot(\beta))$

Answer 2

$sec^2\beta -csc^2\beta \\(1+tan^2\beta )-(1+cot^2\beta )\\1+tan^2\beta -1-cot^2\beta \\tan^2\beta -cot^2\beta \\(tan\beta -cot\beta )(tan\beta +cot\beta )$