Solve this

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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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