MCQ
If $\text{a}\cos\theta+\text{b}\sin\theta=\text{m}$ and $\text{a}\sin\theta-\text{b}\cos\theta=\text{n},$ then $a^2+ b^2=$
- A$m^2-n^2 $
- B$ m^2 n^2 $
- C$ n^2-m^2 $
- ✓$ m^2+n^2 $
✓
Answer
Correct option: D.
$ m^2+n^2 $
$\text{a}\cos\theta+\text{b}\sin\theta=\text{m}$
$\text{a}\sin\theta-\text{b}\cos\theta=\text{n}$
Squaring and adding
$\text{a}^2\cos^2\theta+\text{b}^2\sin^2\theta+2\text{ab }\sin\theta\cos\theta=\text{m}^2$
$\text{a}^2\sin^2\theta+\text{b}^2\cos^2\theta-2\text{ab }\sin\theta\cos\theta=\text{n}^2$
$\text{a}^2(\cos^2\theta+\sin^2\theta)+\text{b}^2(\sin^2\theta+\cos^2\theta)$
$=\text{m}^2+\text{n}^2\ \{\sin^2\theta+\cos^2\theta=1\}$
$\Rightarrow\ \text{a}^2+1+\text{b}^2\times1=\text{m}^2-\text{n}^2$
$\Rightarrow\ \text{a}^2+\text{b}^2=\text{m}^2+\text{n}^2$
Hence, $a^2+ b^2=$
$ m^2+n^2 $
$\text{a}\sin\theta-\text{b}\cos\theta=\text{n}$
Squaring and adding
$\text{a}^2\cos^2\theta+\text{b}^2\sin^2\theta+2\text{ab }\sin\theta\cos\theta=\text{m}^2$
$\text{a}^2\sin^2\theta+\text{b}^2\cos^2\theta-2\text{ab }\sin\theta\cos\theta=\text{n}^2$
$\text{a}^2(\cos^2\theta+\sin^2\theta)+\text{b}^2(\sin^2\theta+\cos^2\theta)$
$=\text{m}^2+\text{n}^2\ \{\sin^2\theta+\cos^2\theta=1\}$
$\Rightarrow\ \text{a}^2+1+\text{b}^2\times1=\text{m}^2-\text{n}^2$
$\Rightarrow\ \text{a}^2+\text{b}^2=\text{m}^2+\text{n}^2$
Hence, $a^2+ b^2=$
$ m^2+n^2 $
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