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Trigonometric Functions — MATHS STD 11 Science — Question

Rajasthan BoardEnglish MediumSTD 11 ScienceMATHSTrigonometric Functions3 Marks
Question
If $\text{a}\cos\theta+\text{b}\sin\theta=\text{m}$ and $\text{a}\sin\theta-\text{b}\cos\theta=\text{n},$ then show that $\text{a}^2+\text{b}^2=\text{m}^2+\text{n}^2.$
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Answer

Given that: $\text{a}\cos\theta+\text{b}\sin\theta=\text{m}$
and $\text{a}\sin\theta-\text{b}\cos\theta=\text{n}$
$\text{R.H.S.}=\text{m}^2+\text{n}^2=(\text{a}\cos\theta+\text{b}\sin\theta)^2+(\text{a}\sin\theta-\text{b}\cos\theta)^2$
$=\text{a}^2\cos^2\theta+\text{b}^2\sin^2\theta+2\text{ab}\sin\theta\cos\theta\\+\text{a}^2\sin^2\theta+\text{b}^2\cos^2\theta-2\text{ab}\sin\theta\cos\theta$
$=\text{a}^2\cos^2\theta+\text{b}^2\sin^2\theta+\text{a}^2\sin^2\theta+\text{b}^2\cos^2\theta$
$=\text{a}^2(\cos^2\theta+\sin^2\theta)+\text{b}^2(\sin^2\theta+\cos^2\theta)$
$=\text{a}^2.1+\text{b}^2.1=\text{a}^2+\text{b}^2\text{ L.H.S.}$
$\text{L.H.S. = R.H.S.}$ Hence proved.

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