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3.Trigonometrical Ratios, Functions and Identities
easy
If $5\tan \theta = 4,$ then $\frac{{5\sin \theta - 3\cos \theta }}{{5\sin \theta + 2\cos \theta }} = $
A
$0$
B
$1$
C
$1/6$
D
$6$
Solution
(c) $5\tan \theta = 4 \Rightarrow \tan \theta = \frac{4}{5}$
$\therefore \sin \theta = \frac{4}{{\sqrt {41} }}$ and
$\cos \theta = \frac{5}{{\sqrt {41} }}$
$\frac{{5\sin \theta – 3\cos \theta }}{{5\sin \theta + 2\cos \theta }}$
$= \frac{{5 \times \frac{4}{{\sqrt {41} }} – 3 \times \frac{5}{{\sqrt {41} }}}}{{5 \times \frac{4}{{\sqrt {41} }} + 2 \times \frac{5}{{\sqrt {41} }}}}$
$\frac{{20 – 15}}{{20 + 10}} = \frac{5}{{30}} = \frac{1}{6}$.
Standard 11
Mathematics