Question

If a cosθ + b sinθ = m and a sinθ – b cosθ = n, then prove that a2 + b2 = m2 + n2​

Answer 1

SOLUTION :-

Given that,

cosθ + b sinθ = m and a sinθ – b cosθ = n .

We have to prove that,

a² + b² = m² + n²

$\longrightarrow\sf (a \ cos\theta+ b\ sin\theta)^2= m^2 \\\\\longrightarrow a^2 \ cos^2\theta+2ab \ cos\theta sin\theta +b^2 \ sin^2\theta= m^2 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ...............(1)$

$\longrightarrow\sf (a \ sin\theta -b \ cos\theta)^2=n^2 \\\\\longrightarrow a^2 \ sin^2\theta-2a \ cos\theta sin\theta+b^2 \ cos^2\theta=n^2 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ..............(2)$

Add equation (1) and equation (2),

$\longrightarrow \sf a^2 \ cos^2\theta + a^2 \ sin^2\theta +b^2 \ sin^2\theta+b^2 \ cos^2 \theta=m^2+n^2 \\\\\longrightarrow a^2(cos^2\theta +sin^2\theta)+b^2(sin^2\theta+cos^2\theta)=m^2+n^2$

We know that,

$\sf sin^2\theta + cos^2\theta = 1$

$\longrightarrow\sf a^2\times 1 +b^2\times 1 = m^2+n^2\\\\\longrightarrow\bf a^2+b^2= m^2+n^2$

Hence proved.