Question

If $4\tan \left(\theta \right)=3$, evaluate $\frac{4\sin \left(\theta \right)-\cos \left(\theta \right)+1}{4\sin \left(\theta \right)+\cos \left(\theta \right)-1}$

Answer 1

Solution:

Given expression is

$\frac{4\sin \left(\theta \right)-\cos \left(\theta \right)+1}{4\sin \left(\theta \right)+\cos \left(\theta \right)-1}$

Dividing and multiplying by $\cos \left(\theta \right)$ in numerator and denominator,

$$\begin{gathered} =\frac{{\displaystyle \frac{4\sin \left(\mathrm{\theta }\right)-\cos \left(\mathrm{\theta }\right)+1}{\cos \left(\mathrm{\theta }\right)}}}{{\displaystyle \frac{4\sin \left(\mathrm{\theta }\right)+\cos \left(\mathrm{\theta }\right)-1}{\cos \left(\mathrm{\theta }\right)}}} \\ =\frac{4{\displaystyle \frac{\sin \left(\mathrm{\theta }\right)}{\cos \left(\mathrm{\theta }\right)}}-{\displaystyle \frac{\cos \left(\mathrm{\theta }\right)}{\cos \left(\mathrm{\theta }\right)}}+{\displaystyle \frac{1}{\cos \left(\mathrm{\theta }\right)}}}{4{\displaystyle \frac{\sin \left(\mathrm{\theta }\right)}{\cos \left(\mathrm{\theta }\right)}}+{\displaystyle \frac{\cos \left(\mathrm{\theta }\right)}{\cos \left(\mathrm{\theta }\right)}}-{\displaystyle \frac{1}{\cos \left(\mathrm{\theta }\right)}}} \end{gathered}$$

$=\frac{4\tan \left(\theta \right)-1+\sec \left(\theta \right)}{4\tan \left(\theta \right)+1-\sec \left(\theta \right)}.........\left(i\right)\left[\because \tan \left(\theta \right)=\frac{\sin \left(\theta \right)}{\cos \left(\theta \right)}\&sec\left(\theta \right)=\frac{1}{\cos \left(\theta \right)}\right]$

Using trigonometric identity: ${\sec }^2\left(\theta \right)=1+{\tan }^2\left(\theta \right)$

$$\begin{gathered} \Rightarrow {\sec }^2\left(\theta \right)=1+{\left(\frac{3}{4}\right)}^2\left[\because 4\tan \left(\theta \right)=3or\tan \left(\theta \right)=\frac{3}{4}\left(Given\right)\right] \\ \Rightarrow {\sec }^2\left(\theta \right)=\frac{25}{16} \\ \Rightarrow \sec \left(\theta \right)=\pm \sqrt{\frac{25}{16}} \\ \Rightarrow \sec \left(\theta \right)=\pm \frac{5}{4} \end{gathered}$$

We will solve for both the values of $\sec \left(\theta \right)=\frac{5}{4}or-\frac{5}{4}$.

Putting values of $\sec \left(\theta \right)=\frac{5}{4}and\tan \left(\theta \right)=\frac{3}{4}$ in expression (i),

$\Rightarrow \frac{4\left({\displaystyle \frac{3}{4}}\right)-1+{\displaystyle \frac{5}{4}}}{4\left({\displaystyle \frac{3}{4}}\right)+{\displaystyle 1}-{\displaystyle \frac{5}{4}}}=\frac{12-4+5}{12+4-5}=\frac{13}{11}$

Putting values of $\sec \left(\theta \right)=-\frac{5}{4}and\tan \left(\theta \right)=\frac{3}{4}$ in (i)

$\Rightarrow \frac{4\left({\displaystyle \frac{3}{4}}\right)-1-{\displaystyle \frac{5}{4}}}{4\left({\displaystyle \frac{3}{4}}\right)+{\displaystyle 1}+{\displaystyle \frac{5}{4}}}=\frac{12-4-5}{12+4+5}=\frac{3}{21}$

Final answer: If$4\tan \left(\theta \right)=3,$value of $\frac{4\sin \left(\theta \right)-\cos \left(\theta \right)+1}{4\sin \left(\theta \right)+\cos \left(\theta \right)-1}=\frac{\mathbf{13}}{\mathbf{11}}or\frac{\mathbf{3}}{\mathbf{21}}$.