Question

tanQ+tan(90°-Q)=sec(90°-Q)​

Answer 1

$\tan(q) + \tan(90 - q) = \sec((90 - q) \\ \\ \tan(q) + \cot(q) = \cos(q) \\ \\ \frac{ \sin(q) }{ \cos(q) } + \frac{ \cos(q) }{ \sin(q) } = \cos(q) \\ \\ \frac{ { \sin(q) }^{2} + { \cos(q) }^{2} }{ \sin(q) \cos(q) } = \cos(q) \\ \\ 1 = { \cos(q) }^{2} \sin(q)$