If a cos θ + b sinθ = 4 and a sin θ – b cos θ = 3, then a² + b² =

We are given the equations:
a cosθ + b sinθ = 4 — (1)
a sinθ – b cosθ = 3 — (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 = 4²
Expand the left-hand side:
a² cos²θ + 2ab cosθ sinθ + b² sin²θ = 16 — (3)

Square both sides of equation (2):
(a sinθ – b cosθ)² = 3²
Expand the left-hand side:
a² sin²θ – 2ab sinθ cosθ + b² cos²θ = 9 — (4)

Step 2: Add equations (3) and (4)
Add the expanded forms of equations (3) and (4):
(a² cos²θ + 2ab cosθ sinθ + b² sin²θ) + (a² sin²θ – 2ab sinθ cosθ + b² cos²θ) = 16 + 9

Simplify the terms:
– The terms involving 2ab cosθ sinθ cancel out.
– Combine the remaining terms:
a² (cos²θ + sin²θ) + b² (sin²θ + cos²θ) = 25

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

Substitute this into the equation:
a² (1) + b² (1) = 25

Simplify:
a² + b² = 25.
The question is connected to Chapter 8, “Introduction to Trigonometry,” in the Class 10th NCERT Mathematics book. Respond based on your comprehension of the chapter.

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