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Introduction of Trigonometry and its Application — Maths STD 10 — Question

CBSE BoardEnglish MediumSTD 10MathsIntroduction of Trigonometry and its Application5 Marks
Question
If $\text{a}\sin\theta+\text{b}\cos\theta=\text{c},$ then prove that $\text{a}\cos\theta-\text{b}\sin\theta=\sqrt{\text{a}^2+\text{b}^2-\text{c}^2}.$
✓

Answer

$\text{a}\sin\theta+\text{b}\cos\theta=\text{c}\ \ [\text{Given}]$
On squaring both sides, we get
$(\text{a}\sin\theta)^2+(\text{b}\cos\theta)^2+2(\text{a}\sin\theta)(\text{b}\cos\theta)=\text{c}^2$
$\Rightarrow\ \text{a}^2\sin^2\theta+\text{b}^2\cos^2\theta+2\text{ab}\sin\theta\cos\theta=\text{c}^2$
$\Rightarrow\ \text{a}^2(1-\cos^2\theta)+\text{b}^2(1-\sin^2\theta)+2\text{ab}\sin\theta\cos\theta=\text{c}^2$
$\Rightarrow\ \text{a}^2-\text{a}^2\cos^2\theta+\text{b}^2-\text{b}^2\sin^2\theta+2\text{ab}\sin\theta\cos\theta=\text{c}^2$
$\Rightarrow\ -\text{a}^2\cos^2\theta-\text{b}^2\sin^2\theta+2\text{ab}\sin\theta\cos\theta=\text{c}^2-\text{a}^2-\text{b}^2$
$\Rightarrow\ \text{a}^2\cos^2\theta+\text{b}^2\sin^2\theta-2\text{ab}\sin\theta\cos\theta=\text{a}^2+\text{b}^2-\text{c}^2$
$\Rightarrow\ (\text{a}\cos\theta)^2+(\text{b}\sin\theta)^2-2(\text{a}\cos\theta)(\text{b}\sin\theta)=\text{a}^2+\text{b}^2-\text{c}^2$
$\Rightarrow\ (\text{a}\cos\theta-\text{b}\sin\theta)^2=\text{a}^2+\text{b}^2-\text{c}^2$
$\Rightarrow\ \text{a}\cos\theta-\text{b}\sin\theta=\pm\sqrt{\text{a}^2+\text{b}^2+\text{c}^2}$
(Taking square root both sides)
Hence, $\text{a}\cos\theta-\text{b}\sin\theta=\sqrt{\text{a}^2+\text{b}^2-\text{c}^2}$
Hence, proved.

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