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Question: If \(a\cos\theta + b\sin\theta = m\) and \(a\sin\theta - b\cos\theta = n,\) then \(a^{2} + b^{2} =\)...

If acos⁡θ+bsin⁡θ=ma\cos\theta + b\sin\theta = m and asin⁡θ−bcos⁡θ=n,a\sin\theta - b\cos\theta = n, then a2+b2=a^{2} + b^{2} =

A

m+nm + n

B

m2−n2m^{2} - n^{2}

C

m2+n2m^{2} + n^{2}

D

None of these

Answer

m2+n2m^{2} + n^{2}

Explanation

Solution

Given that acos⁡θ+bsin⁡θ=ma\cos\theta + b\sin\theta = mand asin⁡θ−bcos⁡θ=n.a\sin\theta - b\cos\theta = n.

Squaring and adding, we get

(acos⁡θ+bsin⁡θ)2+(asin⁡θ−bcos⁡θ)2=m2+n2(a\cos\theta + b\sin\theta)^{2} + (a\sin\theta - b\cos\theta)^{2} = m^{2} + n^{2}

⇒a2(cos⁡2θ+sin⁡2θ)+b2(cos⁡2θ+sin⁡2θ)+2ab(cos⁡θsin⁡θ−sin⁡θcos⁡θ)=m2+n2\Rightarrow a^{2}(\cos^{2}\theta + \sin^{2}\theta) + b^{2}(\cos^{2}\theta + \sin^{2}\theta) + 2ab(\cos\theta\sin\theta - \sin\theta\cos\theta) = m^{2} + n^{2}

Hence, a2+b2=m2+n2.a^{2} + b^{2} = m^{2} + n^{2}.

Trick : Here we can guess that the value of a2+b2a^{2} + b^{2} is independent of θ, so put any suitable value of θ i.e. π2,\frac{\pi}{2}, so that b=mb = m and a=n.a = n. Hence a2+b2=m2+n2.a^{2} + b^{2} = m^{2} + n^{2}.

(Also check for other value of θ).