Question

Hello there!
Prove that :-
tan³ A - 1 / tan A -1 = sec² A + tan A

Answer 1

lhs
${tan}^{3}a - 1 \div tan \: a - 1 \\( tan \: a - 1 )( {tan}^{2}a + 1 + tan \: a) \div tan \: a - 1 \\ {tan}^{2} a + 1 + tan \: a \\ {sec}^{2} a + tan \: a$=rhs
use
a3-b3=(a-b) (a2+ab+b2)
and
sec square a -tan square a =1