MCQ
If $3 \sin x + 4 \cos x = 5,$ then $4 \sin x - 3 \cos x =$
  • A
    $1$
  • B
    $5$
  • C
    $3$
  • $0$

Answer

Correct option: D.
$0$
$\frac{3}{5} \sin x+\frac{4}{5} \cos x=1$
Let $\cos \alpha=\frac{3}{5}$ and $\sin \alpha=\frac{4}{5}$
$\therefore \cos \alpha \sin x+\sin \alpha \cos x=1$
$\Rightarrow \sin (\alpha+x)=\sin \frac{\pi}{2}$
$\Rightarrow \alpha+x=\pi$
$\Rightarrow x=\frac{\pi}{2}-\alpha \ldots . \text { (i) }$
We have to find the value of $4 \sin x - 3 \cos x$
$4 \sin \left(\frac{\pi}{2}-\alpha\right)-3 \cos \left(\frac{\pi}{2}-\alpha\right) \ldots$ From eq. $(i)$
$=4 \cos \alpha-3 \sin \alpha$
$=4 \times \frac{3}{5}-3 \times \frac{4}{5}\left(\because \cos \alpha=\frac{3}{5} \text { and } \sin \alpha=\frac{4}{5}\right)$
$0$

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