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A sin theta -b cos theta=p,a cos theta -b sin theta =q Show that a²+b²=p²+q²​

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Answer:

Explanation:

To show that a² + b² = p² + q², we can start by squaring the given equations and adding them together:

(a sin θ - b cos θ)² + (a cos θ - b sin θ)² = p² + q²

Expanding the squared terms:

(a² sin² θ - 2ab sin θ cos θ + b² cos² θ) + (a² cos² θ - 2ab sin θ cos θ + b² sin² θ) = p² + q²

Combining like terms:

a² sin² θ + a² cos² θ + b² sin² θ + b² cos² θ - 2ab sin θ cos θ - 2ab sin θ cos θ = p² + q²

Using the trigonometric identity sin² θ + cos² θ = 1:

a² + b² - 2ab sin θ cos θ - 2ab sin θ cos θ = p² + q²

Rearranging and factoring out a common factor of 2ab:

(a² + b² - 2ab sin θ cos θ - 2ab sin θ cos θ) = p² + q²

2ab (1 - sin θ cos θ - sin θ cos θ) = p² + q²

2ab (1 - 2 sin θ cos θ) = p² + q²

Using the trigonometric identity 2 sin θ cos θ = sin 2θ:

2ab (1 - sin 2θ) = p² + q²

2ab - 2ab sin 2θ = p² + q²

2ab(1 - sin 2θ) = p² + q²

Since 1 - sin 2θ = cos 2θ:

2ab cos 2θ = p² + q²

Dividing both sides by 2ab:

cos 2θ = (p² + q²) / (2ab)

Using the double angle formula for cosine: cos 2θ = 2cos² θ - 1:

2cos² θ - 1 = (p² + q²) / (2ab)

Rearranging and simplifying:

2cos² θ = (p² + q²) / (2ab) + 1

2cos² θ = (p² + q² + 2ab) / (2ab)

Dividing both sides by 2:

cos² θ = (p² + q² + 2ab) / (4ab)

Using the Pythagorean identity sin² θ + cos² θ = 1, we have:

sin² θ = 1 - cos² θ

sin² θ = 1 - (p² + q² + 2ab) / (4ab)

Rearranging and simplifying:

4ab sin² θ = 4ab - p² - q² - 2ab

4ab sin² θ = 2ab - p² - q²

Multiplying both sides by a² + b²:

4ab (a² + b²) sin² θ = (2ab - p² - q²) (a² + b²)

Expanding both sides:

4a³b sin² θ + 4ab³ sin² θ = 2a³b - p²a² - q²a² - 2ab

User Ajit Kumar Dubey
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