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Trigonometric Identities — Maths STD 10 — Question

Maharashtra BoardEnglish MediumSTD 10MathsTrigonometric Identities4 Marks
Question
Prove the following trigonometric identities.
If $\text{a}\cos\theta+\text{b}\sin\theta=\text{m and a}\sin\theta-\text{b}\cos\theta=\text{n},$ prove that $a^2 + b^2 = m^2 + n^2$.
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Answer

Given,
$\text{a}\cos\theta+\text{b}\sin\theta=\text{m},$
$\text{a}\sin\theta-\text{b}\cos\theta=\text{n}$
We have to prove $a^2 + b^2 = m^2 + n^2$
We know that, $\sin^2\theta+\cos^2\theta=1$
Now, squaring and adding the two equations, we get
$(\text{a}\cos\theta+\text{b}\sin\theta)^2+(\text{a}\sin\theta-\text{b}\cos\theta)^2=\text{m}^2+\text{n}^2$
$\Rightarrow \big(\text{a}^2\cos^2\theta+2\text{ab}\sin\theta\cos\theta+\text{b}^2\sin^2\theta\big)$
$+\big(\text{a}^2\sin^2\theta-2\text{ab}\sin\theta\cos\theta+\text{b}^2\cos^2\theta\big)=2$
$\Rightarrow\ \text{a}^2(\cos^2\theta+\sin^2\theta)+\text{b}^2(\sin^2\theta+\cos^2\theta)=\text{m}^2+\text{n}^2$
$\Rightarrow\ \text{a}^2+\text{b}^2=\text{m}^2+\text{n}^2$
$\therefore\ \text{L.H.S}=\text{R.H.S}$

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