Search any question & find its solution
Question:
Answered & Verified by Expert
If $5 \tan \theta=4$, then $\frac{5 \sin \theta-3 \cos \theta}{5 \sin \theta+2 \cos \theta}=$
Options:
Solution:
2191 Upvotes
Verified Answer
The correct answer is:
$1 / 6$
$\begin{aligned}
5 \tan \theta=4 \Rightarrow \tan \theta & =\frac{4}{5} \\
\therefore \sin \theta & =\frac{4}{\sqrt{41}} \text { and } \cos \theta=\frac{5}{\sqrt{41}} \\
\frac{5 \sin \theta-3 \cos \theta}{5 \sin \theta+2 \cos \theta} & =\frac{5 \times \frac{4}{\sqrt{41}}-3 \times \frac{5}{\sqrt{41}}}{5 \times \frac{4}{\sqrt{41}}+2 \times \frac{5}{\sqrt{41}}} \\
\frac{20-15}{20+10}=\frac{5}{30} & =\frac{1}{6}
\end{aligned}$
5 \tan \theta=4 \Rightarrow \tan \theta & =\frac{4}{5} \\
\therefore \sin \theta & =\frac{4}{\sqrt{41}} \text { and } \cos \theta=\frac{5}{\sqrt{41}} \\
\frac{5 \sin \theta-3 \cos \theta}{5 \sin \theta+2 \cos \theta} & =\frac{5 \times \frac{4}{\sqrt{41}}-3 \times \frac{5}{\sqrt{41}}}{5 \times \frac{4}{\sqrt{41}}+2 \times \frac{5}{\sqrt{41}}} \\
\frac{20-15}{20+10}=\frac{5}{30} & =\frac{1}{6}
\end{aligned}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.