If tanθ =\(\frac{4}{3}\) , then the value of \(\frac{(3sinθ + 2cosθ)}{(3sinθ - 2cosθ)}\) is
Given: tanθ = \(\frac{4}{3}\)
We need to find the value of \(\frac{(3sinθ + 2cosθ)}{(3sinθ - 2cosθ)}\).
We can divide both the numerator and the denominator by cosθ:
\(\frac{(3sinθ + 2cosθ)}{(3sinθ - 2cosθ)} = \frac{3\frac{sinθ}{cosθ} + 2}{3\frac{sinθ}{cosθ} - 2}\)
Since tanθ = \(\frac{sinθ}{cosθ}\), we can substitute this into the expression:
\( \frac{3tanθ + 2}{3tanθ - 2} \)
Now, substitute the given value of tanθ = \(\frac{4}{3}\):
\( \frac{3(\frac{4}{3}) + 2}{3(\frac{4}{3}) - 2} = \frac{4 + 2}{4 - 2} = \frac{6}{2} = 3 \)
Therefore, the value of \(\frac{(3sinθ + 2cosθ)}{(3sinθ - 2cosθ)}\) is 3.
Why other options are incorrect:
- Option 1 (1/2): This is incorrect because it doesn't account for the given value of tanθ correctly.
- Option 2 (1\(\frac{1}{2}\)): This is also an incorrect calculation based on the provided data.
- Option 4 (-3): This would be the result if the signs in the numerator and denominator were reversed; however, the given expression does not have a negative in the numerator.
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