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प्रश्न
If \[3\tan\left( x - 15^\circ \right) = \tan\left( x + 15^\circ \right)\] \[0 < x < 90^\circ\], find θ.
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उत्तर
Given: \[3\tan\left( x - 15^\circ \right) = \tan\left( x + 15^\circ \right)\]
\[\Rightarrow \frac{\tan\left( x + 15^\circ \right)}{\tan\left( x - 15^\circ \right)} = 3\]
Applying componendo and dividendo, we have
\[\frac{\tan\left( x + 15^\circ \right) + \tan\left( x - 15^\circ \right)}{\tan\left( x + 15^\circ \right) - \tan\left( x - 15^\circ \right)} = \frac{3 + 1}{3 - 1}\]
\[ \Rightarrow \frac{\frac{\sin\left( x + 15^\circ \right)}{\cos\left( x + 15^\circ \right)} + \frac{\sin\left( x - 15^\circ \right)}{\cos\left( x - 15^\circ \right)}}{\frac{\sin\left( x + 15^\circ \right)}{\cos\left( x + 15^\circ \right)} - \frac{\sin\left( x - 15^\circ \right)}{\cos\left( x - 15^\circ \right)}} = \frac{4}{2}\]
\[ \Rightarrow \frac{\sin\left( x + 15^\circ \right)\cos\left( x - 15^\circ \right) + \cos\left( x + 15^\circ \right)\sin\left( x - 15^\circ \right)}{\sin\left( x + 15^\circ \right)\cos\left( x - 15^\circ \right) - \cos\left( x + 15^\circ \right)\sin\left( x - 15^\circ \right)} = 2\]
\[ \Rightarrow \frac{\sin\left( x + 15^\circ + x - 15^\circ \right)}{\sin\left( x + 15^\circ- x + 15^\circ \right)} = 2\]
\[\Rightarrow \frac{\sin2x}{\sin30^\circ} = 2\]
\[ \Rightarrow \sin2x = 2 \times \frac{1}{2} = 1 \left( \sin30^\circ = \frac{1}{2} \right)\]
\[ \Rightarrow \sin2x = \sin90^\circ\]
\[ \Rightarrow 2x = 90^\circ \left( 0 < x < 90^\circ \right)\]
\[ \Rightarrow x = 45^\circ\]
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