Question

prove that sec4A-sec2A=tan4A+tan2A

Answer 1

Step-by-step explanation:

Given that:-

LHS:-

Sec^4A-Sec^2A

Sec^2A(Sec^2A-1)

We know that Sec^2A-Tan^2A=1

=>(1+Tan^2A)(Tan^2A)

=>Tan^2A+Tan^4A

=RHS

LHS=RHS

Sec^4A-Sec^2A=Tan^2A+Tan^4A

Used formulae:-

• Sec^2A-Tan^2A=1
• Sec^2A=1+Tan^2A
• Sec^2A-1=Tan^2A

Answer 2

Answer:

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