Question

prove
$\frac{ { \tan(x) }^{3} - 1}{ \tan(x) - 1 } = { \sec(x) }^{2} + \tan(x)$
​

Answer 1

Step-by-step explanation:

tan³x-1/tanx-1

using a³-b³=(a-b)(a²+ab+b²)

(tanx-1)(tan²x+tanx+1)/(tanx-1)

=tan²x+tanx+1

=sec²x+tanx