Question

find the value of

$9 \sec ^{2} (x) - 9 \tan ^{2} (x)$

Answer 1

$9 \sec {}^{2} (x) - \tan {}^{2} (x) \\ \\ = 9( \sec {}^{2} (x) - \tan {}^{2} (x) ) \\ \\ but \: \sec {}^{2} ( \alpha ) - \tan {}^{2} ( \alpha ) = 1 \\ \\ therefore \\ \\ 9( \sec {}^{2} (x) - \tan {}^{2} (x) ) = 9 \times 1 \\ \\ = 9 \\ \\ hope \: it \: helps...$