If a cos θ + b sin θ = m and a sin θ – b cos θ = n, then a² + b² =

We are given the equations:
a cosθ + b sinθ = m — (1)
a sinθ – b cosθ = n — (2)

We need to find the value of a² + b².

Step 1: Square both equations
Square both sides of equation (1):
(a cosθ + b sinθ)² = m²
Expand the left-hand side:
a²cos²θ + 2ab cosθ sinθ + b²sin²θ = m² — (3)

Square both sides of equation (2):
(a sinθ – b cosθ)² = n²
Expand the left-hand side:
a²sin²θ – 2ab sinθ cosθ + b²cos²θ = n² — (4)

Step 2: Add equations (3) and (4)
Add the expanded forms of equations (3) and (4):
(a²cos²θ + 2ab cosθ sinθ + b²sin²θ) + (a²sin²θ – 2ab sinθ cosθ + b²cos²θ) = m² + n²

Simplify the terms:
– The terms involving 2ab cosθ sinθ cancel out.
– Combine the remaining terms:
a²(cos²θ + sin²θ) + b²(sin²θ + cos²θ) = m² + n²

Step 3: Use the Pythagorean identity
From the Pythagorean identity, we know:
cos²θ + sin²θ = 1.

Substitute this into the equation:
a²(1) + b²(1) = m² + n²

Simplify:
a² + b² = m² + n².
The question is based on Chapter 8 of the Class 10th NCERT Mathematics textbook, titled “Introduction to Trigonometry.” Provide your response in line with the concepts covered in this chapter.

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