Question:
If $3 \cot \theta=4$, then $\frac{(5 \sin \theta+3 \cos \theta)}{(5 \sin \theta-3 \cos \theta)}=?$
(a) $\frac{1}{3}$
(b) 3
(c) $\frac{1}{9}$
(d) 9
Solution:
(d) 9
We have $\frac{(5 \sin \theta+3 \cos \theta)}{(5 \sin \theta-3 \cos \theta)}$.
Dividing the numerator and denominator of the given expression by sin θ, we get:
$\frac{\frac{1}{\sin \theta}(5 \sin \theta+3 \cos \theta)}{\frac{1}{\sin \theta}(5 \sin \theta-3 \cos \theta)}$
$=\frac{5+3 \cot \theta}{5-3 \cot \theta}$
$=\frac{5+4}{5-4}=9 \quad[\because 3 \cot \theta=4]$
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