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\begin{array}{l}{\frac{\cos ^{4} \alpha}{\cos ^{2} \beta}+\frac{\sin ^{4} \alpha}{\sin ^{2} \beta}=1} \\ {\cos ^{2} \beta} & {+\frac{\sin ^{4} \beta}{\sin ^{2} \alpha}=1}\end{array} |
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Answer» 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 |
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