Question

prove the following sec^4 a (1-sin^4 a) -2 tan^2 a =1

Answer 1

LHS = sec⁴a (1 - sin⁴a) - 2tan²a

= sec⁴a(1 - sin²a)(1 + sin²a) - 2tan²a

= sec⁴a cos²a (1 + sin²a) - 2tan²a

= sec²a/cos²a × cos²a(1 + sin²a) - 2tan²a [ we know, secA = 1/cosA]

= sec²a (1 + 1 - cos²a) - 2tan²a [ we know , sin²A = 1 - cos²A ]

= sec²a(2 - cos²a) - 2tan²a

= 2sec²a - sec²a.cos²a - 2tan²a

= 2sec²a - 2tan²a - cos²a/cos²a

= 2(sec²a - tan²a) - 1

we know, sec²A - tan²A = 1 so, sec²a - tan²a = 1

= 2 - 1 = 1 = RHS

Answer 2

We need to prove, sec⁴a (1 - sin⁴a) - 2tan²a = 1

Let us take LHS,

sec⁴a (1 - sin⁴a) - 2tan²a

We know,  (1 - sin⁴a) = (1 - sin²a)(1 + sin²a)

Putting above in the given equation, we get

= sec⁴a(1 - sin²a)(1 + sin²a) - 2tan²a

= sec⁴a cos²a (1 + sin²a) - 2tan²a { As we know, (1 - sin²a) = cos²a }

= sec²a/cos²a × cos²a(1 + sin²a) - 2tan²a [ As we know, secA = 1/cosA]

= sec²a (1 + 1 - cos²a) - 2tan²a [ we know , sin²A = 1 - cos²A ]

= sec²a(2 - cos²a) - 2tan²a

= 2sec²a - sec²a.cos²a - 2tan²a

= 2sec²a - 2tan²a - cos²a/cos²a

= 2(sec²a - tan²a) - 1

As we know, sec²A - tan²A = 1

so, sec²a - tan²a = 1

This implies

= 2 - 1 = 1 = RHS

Hence,

sec^4 a (1-sin^4 a) -2 tan^2 a =1