Question

Prove that
(tan A)^2 + (tan A)^4 = (sec A)^4 - (sec A)^2

Answer 1

just use the formula tan²A = sec²A-1
Hope this helps.... :)

Answer 2

LHS

${tan}^{2} a + {tan}^{4} a \\ \\ = {tan}^{2} a(1 + {tan}^{2} a) \\ \\ = {tan}^{2} a {sec}^{2} a \\ \\ = ( {sec}^{2} a - 1) {sec}^{2} a \\ \\ = {sec}^{4} a - {sec}^{2} a$

=RHS