Question

Q19

Q19. If 4 tan $\mathrm{\theta }$ = 3 , evaluate $\left(\frac{4\sin \theta –\cos \theta +1}{4\sin \theta +\cos \theta –1}\right)$

Answer 1

$$\begin{gathered} 4\tan \mathrm{\theta }=3 \\ \Rightarrow \tan \mathrm{\theta }=\frac{3}{4} \\ \Rightarrow \sec \mathrm{\theta }=\sqrt{1+{\tan }^2\mathrm{\theta }}=\sqrt{1+\frac{9}{16}}=\frac{5}{4} \\ \mathrm{Now},\cos \mathrm{\theta }=\frac{1}{\sec \mathrm{\theta }}=\frac{4}{5} \\ \mathrm{Now},\sin \mathrm{\theta }=\sqrt{1-{\cos }^2\mathrm{\theta }}=\sqrt{1-\frac{16}{25}}=\frac{3}{5} \\ \mathrm{Now},\frac{4\sin \mathrm{\theta }-\cos \mathrm{\theta }+1}{4\sin \mathrm{\theta }+\cos \mathrm{\theta }-1} \\ =\frac{4\times {\displaystyle \frac{3}{5}}-{\displaystyle \frac{4}{5}}+1}{4\times {\displaystyle \frac{3}{5}}+{\displaystyle \frac{4}{5}}-1} \\ =\frac{{\displaystyle \frac{8}{5}}+1}{{\displaystyle \frac{16}{5}}-1} \\ =\frac{13}{11} \end{gathered}$$