Question

If p cos θ + q sin θ = u and p sin θ -q cos θ = v; Prove that u2+ v2= p2+q2

Answer 1

Required Answer:-

Given:

• u = p cos θ + q sin θ
• v = p sin θ -q cos θ

To prove:

• u² + v² = p² + q²

Proof:

Taking LHS,

u² + v²

= (p cos θ + q sin θ)² + (p sin θ -q cos θ)²

= p² cos²θ + q² sin²θ + 2pq cosθ sinθ + p² sin²θ + q² cos²θ - 2pq cosθ sinθ

= p² sin²θ + p² cos²θ + q² sin²θ + q² cos²θ

= p²(sin²θ + cos²θ) + q²(sin²θ + cos²θ)

We know that,

➡ sin²θ + cos²θ = 1

So,

p²(sin²θ + cos²θ) + q²(sin²θ + cos²θ)

= p² × 1 + q² × 1

= p² + q²

= RHS (Hence Proved)

Formula used:

• sin²θ + cos²θ = 1 (0° ≤ θ ≤ 90°)

More Formula:

• cosec²θ - cot²θ = 1 (0° ≤ θ ≤ 90°)
• sec²θ - tan²θ = 1 (0° ≤ θ ≤ 90°)
• sin(90° - θ) = cosθ
• cosec(90° - θ) = secθ
• tan(90° - θ) = cotθ