Evaluate  $$\frac{(9sin θ + 2 cos θ) }{ (9sin θ - 2cos θ)}$$​, if 3 tan θ = 4.

Correct option is C

Given:
3tanθ=4
​$$\frac{(9sinθ+2cosθ)}{(9sinθ-2cosθ)}= ?$$​​
​​​Solution:
As per the question,
3tanθ=4
tanθ= $$\frac{4}{3}$$​​
​$$\frac{(9sinθ+2cosθ)}{(9sinθ-2cosθ)}$$

Divide numerator  and denominator by $$cos \theta$$  

​$$\frac{(9tanθ+2)}{(9tanθ-2)}$$ 

= $$\frac{(9\times\frac{4}{3} +2)}{(9\times\frac{4}{3} -2)}$$ 

= $$\frac{14}{10}$$ 

= $$\frac{ 7 }{5}$$​​

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