Question

If 3 sin x + 4 cos x = 5, then 4 sin x − 3 cos x =
(a) 0
(b) 5
(c) 1
(d) None of these

Answer 1

(a) 0

$$\begin{gathered} 3\sin x+4\cos \mathrm{x}=5 \\ \frac{3}{5}\sin x+\frac{4}{5}\cos x=1 \\ \mathrm{Let}\cos \mathrm{\alpha }=\frac{3}{5}\mathrm{and}\sin \mathrm{\alpha }=\frac{4}{5}. \\ \therefore \cos \mathrm{\alpha }\sin x+\mathrm{sin\alpha cos}x=1 \\ \Rightarrow \sin \left(\mathrm{\alpha }+x\right)=\sin \frac{\mathrm{\pi }}{2} \\ \Rightarrow \mathrm{\alpha }+x=\frac{\mathrm{\pi }}{2} \\ \Rightarrow x=\frac{\mathrm{\pi }}{2}-\mathrm{\alpha }....\left(1\right) \\ \mathrm{We}\mathrm{have}\mathrm{to}\mathrm{find}\mathrm{the}\mathrm{value}\mathrm{of}4\sin x-3\cos x. \\ 4\sin \left(\frac{\mathrm{\pi }}{2}-\mathrm{\alpha }\right)-3\cos \left(\frac{\mathrm{\pi }}{2}-\mathrm{\alpha }\right)...\{\mathrm{From}\mathrm{eq}(1\left)\right\} \\ =4\cos \mathrm{\alpha }-3\mathrm{sin\alpha } \\ =4\times \frac{3}{5}-3\times \frac{4}{5}\left(\because \cos \mathrm{\alpha }=\frac{3}{5}\mathrm{and}\sin \mathrm{\alpha }=\frac{4}{5}\right) \\ =0 \end{gathered}$$