If $3 \tan (x - 15 °) = \tan (x + 15 °)$ , $0 < x < 90 °$ , find $θ$ .

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Solution

Given: $3 \tan (x - 15 °) = \tan (x + 15 °)$

$\Rightarrow \frac{\tan (x + 15 °)}{\tan (x - 15 °)} = 3$

Applying componendo and dividendo, we have

$\frac{\tan (x + 15 °) + \tan (x - 15 °)}{\tan (x + 15 °) - \tan (x - 15 °)} = \frac{3 + 1}{3 - 1} \Rightarrow \frac{\frac{\sin (x + 15 °)}{\cos (x + 15 °)} + \frac{\sin (x - 15 °)}{\cos (x - 15 °)}}{\frac{\sin (x + 15 °)}{\cos (x + 15 °)} - \frac{\sin (x - 15 °)}{\cos (x - 15 °)}} = \frac{4}{2} \Rightarrow \frac{\sin (x + 15 °) \cos (x - 15 °) + \cos (x + 15 °) \sin (x - 15 °)}{\sin (x + 15 °) \cos (x - 15 °) - \cos (x + 15 °) \sin (x - 15 °)} = 2 \Rightarrow \frac{\sin (x + 15 ° + x - 15 °)}{\sin (x + 15 ° - x + 15 °)} = 2$

$\Rightarrow \frac{\sin 2 x}{\sin 30 °} = 2 \Rightarrow \sin 2 x = 2 \times \frac{1}{2} = 1 (\sin 30 ° = \frac{1}{2}) \Rightarrow \sin 2 x = \sin 90 ° \Rightarrow 2 x = 90 ° (0 < x < 90 °) \Rightarrow x = 45 °$

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