Question

if cos⁴a/cos²b + sin⁴a/sin²b = 1, then prove that

cos⁴b/cos²a + sin⁴b/sin²a = 1​

Answer 1

Answer:

if cos⁴a/cos²b + sin⁴a/sin²b = 1, then

cos⁴b/cos²a + sin⁴b/sin²a = 1

Answer 2

Question:

If (cos⁴a/cos²b) + (sin⁴a/sin²b) = 1 , Then prove that (cos⁴b/cos²a)+ (sin⁴b+sin²a) = 1

Solution:

Given:

• (cos⁴a/cos²b) + (sin⁴a/sin²b) = 1

To Prove:

• (cos⁴b/cos²a) + (sin⁴b/sin²a) = 1

i) Now let's take:

• (cos⁴A)(cos²B) + (sin⁴A)(sin²B) = 1

Multiplying:

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

We get:

(cos⁴A)/(cos²B) + (sin⁴A)/(sin²B) = (sin²B)(cos²B)

ㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤ↳(1st equation)

ii) Now let's take:

• (cos⁴A)/(cos²B) + (sin⁴B)/(sin²A) = 1

In that let's consider:

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

= (1 - 2sin²A + sin⁴A)(1 - sin²B)

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

ㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤ↳(2nd equation)

Substituting:

• 1st and 2nd equation

We get:

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

⇒ 1 - 2sin²A + sin⁴A - sin²B + 2sin²A × sin²B = sin²B(1 - sin²B) = sin²B - sin⁴B

Rearranging:

We get:

1 - 2(sin²A + sin²B) + (sin⁴A + 2sin²A × sin²B + sin⁴B) = 0

⇒ 1 - 2(sin²A + sin²B) + (sin²A + sin²B)² = 0

Here,

• It is in the form of a² - 2a + 1

• a = (sin²A + sin²B)

⇒ {(sin²A + sin²B) - 1}² = 0

⇒ sin²A + sin²B = 1

ㅤㅤㅤㅤㅤㅤㅤㅤ↳(3rd equation)

From 3rd equation,

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sin²A = 1 - sin²B

⇒ sin²A = cos²B

ㅤㅤㅤㅤㅤㅤㅤ↳(4th equation)

Similarly,

sin²B = 1 - sin²A

⇒ sin²B = cos²A

ㅤㅤㅤㅤㅤㅤㅤ↳(5th equation)

So,

sin⁴B/cos²A = sin⁴B/sin²B

[Substituting: From 5th equation]

⇒ sin⁴B/cos²A = sin²B

ㅤㅤㅤㅤㅤㅤㅤ↳(6th equation)

cos⁴B/sin²A = cos⁴B/cos²B

[Substituting: From 4th equation]

⇒ cos⁴B/sin²A = cos²B

ㅤㅤㅤㅤㅤㅤㅤ↳(7th equation)

Adding: 6th and 7th equation

(sin⁴B/cos²A) + (cos⁴B/sin²A)

We get,

sin²B + sin²B = 1

Hence Verified (sin⁴B/cos²A) + (cos⁴B/sin²A) = 1

$~~$

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