Question

EXERCISE 13B
1. If a cos + b sin 0 = m and a sin 0-bcos O = n, prove that
(m² +n²) = (a² +b²).​

Answer 1

Explanation:

[HERE IT WILL BE Θ INSTEAD OF 0].

Accordingly,

By squaring and adding LHS and RHS of both the equations.

   (a cosΘ + b sinΘ)² + (a sinΘ - b cosΘ)² = m² + n²

⇒ (a²cos²Θ + b²sin²Θ +2ab cosΘsinΘ) + (a²sin²Θ + b²cos²-2abcosΘsinΘ) =m²+n².

⇒ a²(cos²Θ + sin²Θ) +   b²(cos²Θ + sin²Θ) = m²+n²

⇒  a² +  b² = m²+n².                        [By Formula cos²Θ + sin²Θ = 1]

Proved.