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Question

If tanθ =\(\frac{4}{3}\) , then the value of \(\frac{(3sinθ + 2cosθ)}{(3sinθ - 2cosθ)}\) is

This question was previously asked in
SSC CGL 2016 (Tier 1) Previous Year Question Paper (11-Sep-2016) (Shift 2)
The correct answer is
3

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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