If cot⁴φ − cot²φ = 1, then find the value of cos⁴φ + cos²φ.

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Accepted Answer

Given: cot⁴φ − cot²φ = 1

Step 1: Let x = cot²φ.
Then the equation becomes:

x² - x = 1

Step 2: Rearranging:

x² - x - 1 = 0

Step 3: Solve the quadratic equation for x:

x = [1 ± √(1 + 4)] / 2 = [1 ± √5] / 2

Since cot²φ is positive, take the positive root:

cot²φ = (1 + √5) / 2

Step 4: Recall that cot²φ = cos²φ / sin²φ.

We want to find cos⁴φ + cos²φ.

Step 5: Use the Pythagorean identity:
sin²φ = 1 - cos²φ

Step 6: Express cot²φ in terms of cos²φ:

cot²φ = cos²φ / (1 - cos²φ) = (1 + √5) / 2

Step 7: Let y = cos²φ.

Then:

y / (1 - y) = (1 + √5) / 2

Cross-multiplied:

2y = (1 + √5)(1 - y)

2y = (1 + √5) - (1 + √5)y

Bring terms involving y to one side:

2y + (1 + √5)y = 1 + √5

y(2 + 1 + √5) = 1 + √5

y(3 + √5) = 1 + √5

Step 8: Solve for y:

y = (1 + √5) / (3 + √5)

Step 9: Rationalize the denominator:

Multiply numerator and denominator by (3 - √5):

y = [(1 + √5)(3 - √5)] / [(3 + √5)(3 - √5)]

Calculate numerator:

(1)(3) + (1)(-√5) + (√5)(3) + (√5)(-√5) = 3 - √5 + 3√5 - 5 = (3 - 5) + (-√5 + 3√5) = -2 + 2√5

Calculate denominator:

3² - (√5)² = 9 - 5 = 4

So,

y = (-2 + 2√5) / 4 = (2√5 - 2) / 4 = (√5 - 1) / 2

Step 10: Now find cos⁴φ + cos²φ = y² + y

Calculate y²:

y² = [(√5 - 1)/2]² = ( (√5 - 1)² ) / 4 = (5 - 2√5 + 1) / 4 = (6 - 2√5) / 4 = (3 - √5) / 2

Step 11: Sum y² + y:

= (3 - √5)/2 + (√5 - 1)/2 = [ (3 - √5) + (√5 - 1) ] / 2 = (3 - 1) / 2 = 2 / 2 = 1

Step 12: Re-check the arithmetic in step 11:

(3 - √5) + (√5 - 1) = 3 - √5 + √5 - 1 = 3 - 1 = 2

Divide by 2:

2 / 2 = 1

So, cos⁴φ + cos²φ = 1

Step 13: But this contradicts the earlier option marked correct. Let's verify carefully.

Re-examining step 10 and 11:

cos⁴φ + cos²φ = y² + y

We have y = (√5 - 1)/2

Calculate y²:

(√5 - 1)² = 5 - 2√5 + 1 = 6 - 2√5

So y² = (6 - 2√5)/4 = (3 - √5)/2

Sum y² + y = (3 - √5)/2 + (√5 - 1)/2 = (3 - √5 + √5 - 1)/2 = (3 - 1)/2 = 2/2 = 1

Therefore, the value is 1.

Step 14: Final answer is 1.

Hence, option 1 is correct.

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