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Question
If 3 tan (θ – 15°) = tan (θ + 15°), 0° < θ < 90°, then θ = ______.
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Solution
If 3 tan (θ – 15°) = tan (θ + 15°), 0° < θ < 90°, then θ = `pi/4`.
Explanation:
Given that 3 tan (θ – 15°) = tan (θ + 15°)
Which can be rewritten as `(tan(theta + 15^circ))/(tan(theta - 15^circ)) = 3/1`
Applying componendo and Dividendo
We get `(tan(theta + 15^circ) + tan(theta - 15^circ))/(tan(theta + 15^circ) - tan(theta - 15^circ))` = 2
⇒ `(sin(theta + 15^circ) cos(theta - 15^circ) + sin(theta - 15^circ) cos(theta + 15^circ))/(sin(theta + 15^circ) cos(theta - 15^circ) - sin(theta - 15^circ) cos(theta + 15^circ))` = 2
⇒ `(sin 2theta)/(sin30^circ)` = 2
i.e., sin 2θ = 1
Giving θ = `pi/4`
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