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`        If (cos⁴a/cos²b) + (sin⁴a/sin²b) = 1Then prove that......(cos⁴b/cos²a)+ (sin⁴b+sin²a)= 1`
3 years ago

Nishant Vora
IIT Patna
2467 Points
```							Dear student,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*sin²Biii) 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 - sin⁴BRearranging, 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 = 1Thus it is proved that, (sin⁴B/cos²A) + (cos⁴B/sin²A) = 1
```
3 years ago
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### Course Features

• 31 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions