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Answer 1

$\blue{\bold{\underline{\underline{Answer:}}}}$

$\green{\tt{\therefore{tan \: \frac{ \theta_{1} - \theta_{2} }{2} \: tan \: \frac{ \theta_{1} + \theta_{2} }{2} = - \frac{1}{3}}}}$

$\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\green{\underline \bold{Given :}} \\ \tt: \implies cos \: \theta_{1} = 2 \: cos \: \theta_{2} \\ \\ \red{\underline \bold{To \: Find :}} \\ \tt: \implies tan \: \frac{ \theta_{1} - \theta_{2} }{2} \: tan \: \frac{ \theta_{1} + \theta_{2} }{2} =?$

• According to given question :

$\bold{As \: we \: know \: that} \\ \tt: \implies tan \: \frac{ \theta_{1} - \theta_{2} }{2} \: tan \: \frac{ \theta_{1} + \theta_{2} }{2} \\ \\ \tt: \implies \frac{ sin \: \frac{\theta_{1} - \theta_{2}}{2} }{cos \: \frac{\theta_{1} - \theta_{2}}{2} } \times \frac{ sin \: \frac{\theta_{1} + \theta_{2}}{2} }{cos \: \frac{\theta_{1} + \theta_{2}}{2} } \\ \\ \text{Multiplying \: both \: in \: numerator \: and \: denominator \: by \: 2} \\ \\ \tt: \implies \frac{ 2 \: sin \: \frac{\theta_{1} - \theta_{2}}{2} }{2 \: cos \: \frac{\theta_{1} - \theta_{2}}{2} } \frac{ sin \: \frac{\theta_{1} + \theta_{2}}{2} }{cos \: \frac{\theta_{1} + \theta_{2}}{2} } \\ \\ \tt \circ \: 2 \: sin \: A\: sin \:B = cos(A - B) - cos(A+ B)$

$\tt \circ \: 2 \: cos \: A\: cos\: B= cos(A + B) + cos(A - B) \\ \\ \tt: \implies \frac{cos \: \bigg(\frac{\theta_{1} - \theta_{2}}{2} - (\frac{\theta_{1} + \theta_{2}}{2} ) \bigg) - cos \: \bigg(\frac{\theta_{1} - \theta_{2}}{2} + \frac{\theta_{1} + \theta_{2}}{2} \bigg)}{cos \: \bigg(\frac{\theta_{1} - \theta_{2}}{2} + (\frac{\theta_{1} + \theta_{2}}{2} ) \bigg) + cos \: \bigg(\frac{\theta_{1} - \theta_{2}}{2} - (\frac{\theta_{1} + \theta_{2}}{2} ) \bigg)} \\ \\ \tt: \implies \frac{cos( - \theta_{2}) - cos \: \theta_{1} }{cos \: \theta_{1} + cos( - \theta_{2})} \\ \\ \tt \circ \: cos( - \theta) = cos \: \theta \\ \\ \tt: \implies \frac{cos \: \theta_{2} - cos \: \theta_{1}}{cos \: \theta_{1} + cos \: \theta_{2}} \\ \\ \tt: \implies \frac{cos \: \theta_{2} - 2cos \: \theta_{2}}{2cos \: \theta_{2} +cos \: \theta_{2}} \\ \\ \tt: \implies - \frac{cos \: \theta_{2}}{3cos \: \theta_{2}} \\ \\ \green{\tt: \implies tan \: \frac{ \theta_{1} - \theta_{2} }{2} \: tan \: \frac{ \theta_{1} + \theta_{2} }{2} = - \frac{1}{3} }$