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$m^2n^2$
- 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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