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If cos4αcos2β+sin4αsin2β = 1, prove that cos4βcos2α+sin4βsin2α = 1 - Mathematics

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प्रश्न

If `(cos^4alpha)/(cos^2beta) + (sin^4alpha)/(sin^2beta)` = 1, prove that `(cos^4beta)/(cos^2alpha) + (sin^4beta)/(sin^2alpha)` = 1

बेरीज
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उत्तर

`(cos^4beta)/(cos^2alpha) + (sin^4beta)/(sin^2alpha) =  (cos^2beta*cos^2beta)/(cos^2alpha) + (sin^2beta*sin^2beta)/(sin^2alpha)`  ......(1)

Given `(cos^4alpha)/(cos^2beta) + (sin^4alpha)/(sin^2beta)` = 1

`(cos^4alpha*sin^2beta + sin^4alpha*cos^2beta)/(cos^2beta sin^2beta)` = 1

`cos^4alpha sin^2beta + sin^4alpha*cos^2beta = cos^2beta sin^2beta`

`cos^4alpha(1 - cos^2beta) + (1 - cos^2alpha)^2 cos^2beta = cos^2beta(1 - cos^2beta)`

`cos^4alpha - cos^4alpha cos^2beta + (1 - 2cos^2alpha + cos^4alpha) cos^2beta = cos^2beta - cos^4beta`

`cos^4alpha - cos^4alpha cos^2beta + cos^2beta - 2cos^2alpha cos^2beta + cos^4alpha cos^2beta = cos^2beta - cos^4beta`

`cos^4alpha - 2cos^2alpha cos^2beta + cos^4beta` = 0

`(cos^2alpha - cos^2beta)^2` = 0

`cos^2alpha = cos^2beta`  ......(2)

`1 - sin^2alpha = 1 - sin^2beta`

`sin^2alpha = sin^2beta`  ......(3

Using equation (2) and (3), equation (1) becomes

`(cos^4beta)/(cos^2alpha) + (sin^4beta)/(sin^2alpha) = (cos^2beta*cos^2beta)/(cos^2alpha) + (sin^2beta*sin^2beta)/(sin^2alpha)`

= `(cos^2beta*cos^2alpha)/(cos^2alpha) + (sin^2beta*sin^2alpha)/(sin^2alpha)`

= `cos^2beta + sin^2beta`

= 1

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A Recall of Basic Results
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 3: Trigonometry - Exercise 3.1 [पृष्ठ ९२]

APPEARS IN

सामाचीर कलवी Mathematics - Volume 1 and 2 [English] Class 11 TN Board
पाठ 3 Trigonometry
Exercise 3.1 | Q 5.. (ii) | पृष्ठ ९२

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