Question

$\mathrm{If}\sin \theta =\frac{4}{\sqrt{41}}\mathrm{where}\frac{\pi }{2}<\theta <\pi ,\mathrm{then}\frac{5\sin \theta -3\cos \theta }{5\sin \theta +3\cos \theta }\mathrm{is}\mathrm{equal}\mathrm{to}$

• A. 5
• B. 6
• C. 7
• D. 8

Answer 1

The correct option is C 7

For solving this type of questions, we need to find the quadrant of $\theta$.

Here, $\frac{\pi }{2}<\theta <\pi$ means the $\theta$ is in 2nd quadrant.
Now, $\sin \theta =\frac{4}{\sqrt{41}}$$\Rightarrow \cos \theta =\sqrt{1-{\sin }^2\theta }\Rightarrow \cos \theta =\sqrt{1-\frac{16}{41}}\Rightarrow \cos \theta =\pm \frac{5}{\sqrt{41}}$
Since $\theta$ lies in 2nd quadrant, $\cos \theta$ is negative.
$\Rightarrow \cos \theta =-\frac{5}{\sqrt{41}}$.
$\Rightarrow \frac{5\sin \theta -3\cos \theta }{5\sin \theta +3\cos \theta }=\frac{5\times \frac{4}{\sqrt{41}}-3\times \frac{-5}{\sqrt{41}}}{5\times \frac{4}{\sqrt{41}}+3\times \frac{-5}{\sqrt{41}}}=\frac{20+15}{20-15}=7$
Hence, 7 is the correct answer.