Question

If (cos⁴A/cos²B) + (sin⁴A/sin²B) = 1
Prove that
(cos⁴B/cos²A) + (sin⁴B/sin²A) = 1

Answer 1

here is the solution to your question.

i) Multiplying by (sin²B)(cos²B), the given expression changes as:

(cos⁴A)(cos²B) + (sin⁴A)(sin²B) = (sin²B)(cos²B) -------- (1)

ii) (cos⁴A)(cos²B) = {(1 - sin²A)²}*(1 - sin²B)

= (1 - 2sin²A + sin⁴A)(1 - sin²B)

= 1 - 2sin²A + sin⁴A - sin²B + 2sin²A*sin²B - sin⁴A*sin²B

iii) Substituting this in (1) above,

1 - 2sin²A + sin⁴A - sin²B + 2sin²A*sin²B - sin⁴A*sin²B + sin⁴A*sin²B = sin²B*cos²B

==> 1 - 2sin²A + sin⁴A - sin²B + 2sin²A*sin²B = sin²B(1 - sin²B) = sin²B - sin⁴B

Rearranging, 1 - 2(sin²A + sin²B) + (sin⁴A + 2sin²A*sin²B + sin⁴B) = 0

==> 1 - 2(sin²A + sin²B) + (sin²A + sin²B)² = 0
[This is in the form of a² - 2a + 1, where a = (sin²A + sin²B)]

==> {(sin²A + sin²B) - 1}² = 0

==> sin²A + sin²B = 1 ---- (2)

iv) From (2) above, sin²A = 1 - sin²B; ==> sin²A = cos²B ------- (3)
Similarly, sin²B = 1 - sin²A; ==> sin²B = cos²A -------- (4)

v) So, sin⁴B/cos²A = sin⁴B/sin²B [Substituting from (4) above]
==> sin⁴B/cos²A = sin²B -------- (5)

cos⁴B/sin²A = cos⁴B/cos²B [Substituting from (3) above]
==> cos⁴B/sin²A = cos²B -------- (6)

Adding (5) & (6),

(sin⁴B/cos²A) + (cos⁴B/sin²A) = sin²B + sin²B = 1

Thus it is proved that, (sin⁴B/cos²A) + (cos⁴B/sin²A) = 1