If \( \tan \theta = \frac{3}{4} \) and \( \theta \) is in the first quadrant, find the value of \( \sin \theta + \cos \theta \).
Show Hint
- \( \frac{5}{4} \)
- \( \frac{7}{5} \)
- \( \frac{5}{7} \)
\( \frac{4}{5} \)
The Correct Option is B
Solution and Explanation
Given \( \tan \theta = \frac{3}{4} \) in the first quadrant, where \( \sin \theta \) and \( \cos \theta \) are positive. Since \( \tan \theta = \frac{\sin \theta}{\cos \theta} \), assume a right triangle with opposite side 3 and adjacent side 4. The hypotenuse is: \[ \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \] Thus: \[ \sin \theta = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{3}{5}, \quad \cos \theta = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{4}{5} \] Calculate: \[ \sin \theta + \cos \theta = \frac{3}{5} + \frac{4}{5} = \frac{7}{5} \] Alternatively, use the identity: \[ \sin \theta + \cos \theta = \sqrt{2} \sqrt{\sin^2 \theta + \cos^2 \theta + 2 \sin \theta \cos \theta} = \sqrt{2} \sqrt{1 + 2 \cdot \frac{\tan \theta}{1 + \tan^2 \theta}} \] Since \( \tan \theta = \frac{3}{4} \), this is complex, so the triangle method is simpler. The value is: \[ {\frac{7}{5}} \]
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