Question

If $5\tan \theta =4,$ then $\frac{(5\sin \theta -3\cos \theta )}{(5\sin \theta +2\cos \theta )}$ is

Answer 1

Finding the value of $\frac{\mathbf{(}\mathbf{5}\mathbf{sin}\mathbf{\theta }\mathbf{-}\mathbf{3}\mathbf{cos}\mathbf{\theta }\mathbf{)}}{\mathbf{(}\mathbf{5}\mathbf{sin\theta }\mathbf{+}\mathbf{2}\mathbf{cos\theta }\mathbf{)}}$ :

Given that

$$\begin{gathered} 5\tan \theta =4, \\ \Rightarrow \tan \theta =\frac{4}{5} \end{gathered}$$

Now ,$\frac{(5\sin \theta -3\cos \theta )}{(5\sin \theta +2\cos \theta )}$

Dividing the numerator and denominator by $\cos \theta$

$$\begin{gathered} =\frac{(5\tan \theta –3)}{(5\tan \theta +2)} \\ =\frac{\left[5\right({\displaystyle \frac{4}{5}})–3]}{\left[5\right({\displaystyle \frac{4}{5}})+2]} \\ =\frac{(4–3)}{(4+2)} \\ =\frac{1}{6} \end{gathered}$$

Hence the value of $\frac{\mathbf{(}\mathbf{5}\mathbf{s}\mathbf{i}\mathbf{n}\mathbf{\theta }\mathbf{-}\mathbf{3}\mathbf{c}\mathbf{o}\mathbf{s}\mathbf{\theta }\mathbf{)}}{\mathbf{(}\mathbf{5}\mathbf{s}\mathbf{i}\mathbf{n}\mathbf{\theta }\mathbf{+}\mathbf{2}\mathbf{c}\mathbf{o}\mathbf{s}\mathbf{\theta }\mathbf{)}}$ is $\frac{\mathbf{1}}{\mathbf{6}}$.