Question

If a coso +b sino =m and a sino-b coso=n,prove that a^2+b^2=m^2+n^2​

Answer 1

Answer:

a²+b² = m²+n²

Given :-

a Cos θ + b Sin θ = m

a Sin θ - b Cos θ = n

To find :-

Prove that a²+b² = m²+n²

Solution :-

Given that :-

a Cos θ + b Sin θ = m

On squaring both sides then

=> (a Cos θ + b Sin θ)² = m²

=>a²Cos²θ+2abCosθSinθ+b²Sin²θ=m²---(1)

Since (a+b)²=a²+2ab+b²

and

a Sin θ - b Cos θ = n

On squaring both sides then

=> (a Sin θ - b Cos θ)² = n²

=>a²Sin²θ-2abSinθCosθ+b²Cos²θ=n²---(2)

Since (a-b)²=a²-2ab+b²

On adding (1)&(2)

a²Cos²θ+2abCosθSinθ+b²Sin²θ=m²

a²Sin²θ-2abSinθCosθ+b²Cos²θ=n²

(+)

______________________________

a²Cos²θ+a²Sin²θ+b²Sin²θ+b²Cos²θ

=m²+n²

______________________________

=> a²Cos²θ+a²Sin²θ+b²Sin²θ+b²Cos²θ

=m²+n²

=> a²(Cos²θ+Sin²θ)+b²(Sin²θ+Cos²θ)

=m²+n²

We know that Sin²A + Cos² A = 1

=> a²(1)+b²(1) = m²+n²

=> a²+b² = m²+n²

Hence, Proved.

Used formulae:-

• (a-b)²=a²-2ab+b²

• (a+b)²=a²+2ab+b²

• Sin²A + Cos² A = 1