Please prove that sec^(2) \beta - cosec^(2) = (tan \beta + cot \beta) * (tan \beta - cot \beta)

Question

Please prove that


`sec^(2) \beta - cosec^(2) = (tan \beta + cot \beta) * (tan \beta - cot \beta)`

Answer 1

`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 )`

Answer 2

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))`