Question

If $\sin \theta +2\cos \theta =1$ prove that $2\sin \theta -\cos \theta =2$.

Answer 1

It is given that,
$$\begin{gathered} \sin \theta +2\cos \theta =1 \\ \Rightarrow 2\cos \theta =1-\sin \theta .....\left(\mathrm{i}\right) \\ \mathrm{Squaring}\mathrm{both}\mathrm{sides},\mathrm{we}\mathrm{get} \\ \Rightarrow {\left(2\cos \theta \right)}^2={\left(1-\sin \theta \right)}^2 \\ \Rightarrow 4{\cos }^2\theta =1+{\sin }^2\theta -2\sin \theta \\ \Rightarrow 4{\cos }^2\theta =1+{\sin }^2\theta -2\sin \theta -1+1 \\ \Rightarrow 4{\cos }^2\theta =2-2\sin \theta -\left(1-{\sin }^2\theta \right) \\ \Rightarrow 4{\cos }^2\theta =2-2\sin \theta -{\cos }^2\theta \\ \Rightarrow 5{\cos }^2\theta =2\left(1-\mathrm{sin\theta }\right) \\ \Rightarrow 5{\cos }^2\theta =4\cos \theta \left[\mathrm{Using}\left(\mathrm{i}\right)\right] \\ \Rightarrow \cos \theta \left(5\cos \theta -4\right)=0 \\ \Rightarrow \cos \theta =0,\cos \theta =\frac{4}{5} \\ \mathrm{Putting}\cos \theta =\frac{4}{5}\mathrm{in}\sin \theta +2\cos \theta =1,\mathrm{we}\mathrm{get} \\ \Rightarrow \sin \theta =1-2\cos \theta =1-2\left(\frac{4}{5}\right)=-\frac{3}{5} \\ \mathrm{This}\mathrm{is}\mathrm{not}\mathrm{possible}. \\ \mathrm{Putting}\cos \theta =0\mathrm{in}\sin \theta +2\cos \theta =1,\mathrm{we}\mathrm{get} \\ \mathrm{sin\theta }=1 \\ \mathrm{Thus},\mathrm{the}\mathrm{value}\mathrm{of}2\sin \theta -\cos \theta =2\left(1\right)-0=2. \end{gathered}$$