Question

$\mathrm{If}3\cot \theta =4,\mathrm{then}\mathrm{write}\mathrm{the}\mathrm{value}\mathrm{of}\frac{\left(2\cos \theta +\sin \theta \right)}{\left(4\cos \theta -\sin \theta \right)}.$

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ 3\cot \theta =4 \\ \Rightarrow \cot \theta =\frac{4}{3} \\ \mathrm{Now}, \\ \frac{\left(2\cos \theta +\sin \theta \right)}{\left(4\cos \theta -\sin \theta \right)} \\ =\frac{\left({\displaystyle \frac{2\cos \theta }{\sin \theta }}+{\displaystyle \frac{\sin \theta }{\sin \theta }}\right)}{\left({\displaystyle \frac{4\cos \theta }{\sin \theta }}-{\displaystyle \frac{\sin \theta }{\sin \theta }}\right)}\left(\mathrm{Dividing}\mathrm{numerator}\mathrm{and}\mathrm{denominator}\mathrm{by}\sin \theta \right) \\ =\frac{\left(2\cot \theta +1\right)}{\left(4\cot \theta -1\right)} \\ =\frac{\left(2\times {\displaystyle \frac{4}{3}}+1\right)}{\left(4\times {\displaystyle \frac{4}{3}}-1\right)} \\ =\frac{\left({\displaystyle \frac{8}{3}}+{\displaystyle \frac{1}{1}}\right)}{\left({\displaystyle \frac{16}{3}}-{\displaystyle \frac{1}{1}}\right)} \\ =\frac{\left({\displaystyle \frac{8+3}{3}}\right)}{\left({\displaystyle \frac{16-3}{3}}\right)} \\ =\frac{\left({\displaystyle \frac{11}{3}}\right)}{\left({\displaystyle \frac{13}{3}}\right)} \\ =\frac{11}{13} \end{gathered}$$