Question

a*cos o + b*sin o=m
a*sin o + b*cosec o=n
to prove....m² +n²= a²+b²​

Answer 1

Step-by-step explanation:

given, a.cosθ+b.sinθ = m

and a.sinθ - b.cosθ = n (please check your question)

consider, m² + n²

=> (a.cosθ+b.sinθ)² + (a.sinθ - b.cos-θ)²

=> (a².cos²θ+b².sin²θ +2a.cosθ.b.sinθ) + (a².sin²θ + b².cos²θ-2a.sinθ.b.cosθ)

=> a².cos²θ+b².sin²θ + a².sin²θ + b².cos²θ

=> a²(sin²θ+cos²θ) + b²(sin²θ+cos²θ)

=> a²+ b²

ie, m² + n² = a²+ b²

Hence proved