Question

Show that tan^4A+tan^2A=sec^4A-sec^2A

Answer 1

Given, RHS = sec^4A-sec^2A

                    = sec^2A ( sec^2A - 1)

                    = sec^2A tan^2A

                    = (tan^2A + 1) tan^2A

                    = tan^4A + tan^2A.

LHS = RHS.


Hope this helps!

Answer 2

Answer:

LHS

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