Question

sec^4 x-tan^4x=1+2tan^2x

Answer 1

Given Equation is sec^4x - tan^4x.

We know that a^2 - b^2 = (a+b)(a-b).

= (sec^2x + tan^2x)(sec^2x-tan^2x)

= (1/cos^2x + sin^2x/cos^2x)(1/cos^2x - sin^2x/cos^2x)

= (1 + sin^2x/cos^2x)(1 - sin^2x/cos^2x)

We know that 1 - sin^2 theta = cos^2 theta

= (1 + sin^2x/cos^2x)(cos^2x/cos^2x)

= (1 + sin^2x/cos^2x) * 1

= (1 + sin^2x/cos^2x)

= (1/cos^2x + sin^2x/cos^2x)

= sec^2x + tan^2x

We know that 1 + tan^2 theta = sec^2 theta

= 1 + tan^2 x + tan^2x

= 1 + 2 tan^2x.


Hope this helps!

Answer 2

LHS
1+tan^2x=sec^2x
1= sec^2x-tan^2x
formula a^2-b^2=(a+b)(a-b)
sec^4x-tan^4x=(sec^2x+tan^2x)(sec^2x-tan^2x)
= sec^2x + tan^2x
= 1+tan^2x+tan^2x
= 1+2tan^2x= RHS