please answer this quality answer will be marked as brainliest
if spammed immediately reported
explain briefly
Share
please answer this quality answer will be marked as brainliest
if spammed immediately reported
explain briefly
Sign Up to our social questions and Answers Engine to ask questions, answer people’s questions, and connect with other people.
Login to our social questions & Answers Engine to ask questions answer people’s questions & connect with other people.
Verified answer
[tex]\blue{\bold{\underline{\underline{Answer:}}}}[/tex]
[tex]\green{\tt{\therefore{tan \frac{ \theta + \alpha }{2} tan \frac{ \theta - \alpha }{2} = {tan}^{2} \: \frac{ \beta }{2}}}}\\[/tex]
[tex]\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}[/tex]
[tex] \green{\underline \bold{Given :}} \\ \tt: \implies cos \: \theta = cos \: \alpha \: cos \: \beta \\ \\ \red{\underline \bold{To \: Find :}} \\ \tt: \implies tan \frac{ \theta + \alpha }{2} tan \frac{ \theta - \alpha }{2} = ? [/tex]
• According to given question :
[tex] \bold{As \: we \: know \: that} \\ \tt: \implies\tan \frac{ \theta + \alpha }{2} tan \frac{ \theta - \alpha }{2} \\ \\ \tt: \implies \frac{sin \: \bigg (\frac{ \theta + \alpha }{2} \bigg)}{cos \bigg( \frac{ \theta + \alpha }{2} \bigg)} \times \frac{sin \bigg(\frac{ \theta - \alpha }{2} \: \bigg)}{cos \: \bigg(\frac{ \theta - \alpha }{2} \bigg)} \\ \\ \text{Multiplying \: both \: in \: numerator \: and \: denominator \:by \: 2} \\ \\ \tt: \implies \frac{2 \: sin \: \bigg (\frac{ \theta + \alpha }{2} \bigg)}{2 \: cos \bigg( \frac{ \theta + \alpha }{2} \bigg)} \frac{sin \bigg(\frac{ \theta - \alpha }{2} \: \bigg)}{cos \: \bigg(\frac{ \theta - \alpha }{2} \bigg)} [/tex]
[tex] \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 + \alpha }{2} - \bigg( \frac{ \theta - \alpha }{2} \bigg) \bigg) - cos \bigg( \frac{ \theta + \alpha }{2} + \frac{ \theta - \alpha }{2} \bigg ) }{cos \bigg( \frac{ \theta + \alpha }{2} + \frac{ \theta - \alpha }{2} \bigg) + cos \bigg( \frac{ \theta + \alpha }{2} - \bigg( \frac{ \theta - \alpha }{2} \bigg)\bigg )} \\ \\ \tt: \implies \frac{cos \bigg( \frac{ \theta + \alpha - \theta + \alpha }{2} \bigg) - cos \bigg( \frac{ \theta + \alpha }{2} + \frac{ \theta - \alpha }{2} \bigg ) }{cos \bigg( \frac{ \theta + \alpha }{2} + \frac{ \theta - \alpha }{2} \bigg) + cos \bigg( \frac{ \theta + \alpha - \theta + \alpha }{2} \bigg )} [/tex]
[tex] \tt: \implies \frac{cos \: \alpha - cos \: \theta}{cos \: \theta + cos \: \alpha } \\ \\ \tt: \implies \frac{cos \: \alpha - cos \: \alpha \:cos \: \beta }{cos \: \alpha \: cos \: \beta + cos \: \alpha } \\ \\ \tt: \implies \frac{cos \alpha (1 - cos \: \beta) }{cos \: \alpha (cos \: \beta + 1)} \\ \\ \tt: \implies \frac{1 - cos \: \beta }{cos \: \beta + 1} \\ \\ \tt \circ \: cos \: 2 x = 1 - 2{sin}^{2} \:x \\ \\ \tt \circ \: 2x = \beta \implies x = \frac{ \beta }{2} \\ \\ \tt \circ \:cos \: \beta = 1 - 2 {sin}^{2} \: \frac{ \beta }{2} \\ \\ \tt: \implies \frac{1 - \bigg(1 - 2 {sin}^{2} \: \frac{ \beta }{2} \bigg) }{cos \: \beta + 1} \\ \\ \tt: \implies \frac{2 {sin}^{2} \: \frac{ \beta }{2} }{cos \: \beta + 1} [/tex]
[tex] \tt \circ \: cos \: 2 x = 2 cos^{2} \: x - 1 \\ \\ \tt \circ \: cos \: \beta = 2 {cos}^{2} \: \frac{ \beta }{2} - 1 \\ \\ \tt: \implies \frac{2sin^{2} \: \frac{ \beta }{2} }{2cos^{2} \: \frac{ \beta }{2} } \\ \\ \tt: \implies \frac{sin^{2} }{cos^{2} } \bigg (\frac{ \beta }{2} \bigg) \\ \\ \green{\tt : \implies {tan}^{2} \: \frac{ \beta }{2} } \\ \\ \green{\tt \therefore tan \frac{ \theta + \alpha }{2} tan \frac{ \theta - \alpha }{2} = {tan}^{2} \: \frac{ \beta }{2} }[/tex]
we have,
[tex] \tan \frac{\theta + \alpha }{2} \tan \frac{\theta - \alpha }{2} = \big( \frac{ \tan \frac{ \theta}{2} + \tan \frac{ \alpha }{2} }{1 - \tan \frac{\theta}{2} \tan \frac{ \alpha }{2} } \big)\big( \frac{ \tan \frac{ \theta}{2} - \tan \frac{ \alpha }{2} }{1 + \tan \frac{\theta}{2} \tan \frac{ \alpha }{2} } \big)[/tex]
[tex] = \frac{ { \tan }^{2} \frac{\theta}{2} - { \tan }^{2} \frac{ \alpha }{2} }{1 - { \tan }^{2} \frac{\theta}{2} { \tan}^{2} \theta } [/tex]
[tex] = \frac{ \frac{(1 - \cos\theta) }{(1 + \cos\theta) } - \frac{(1 - \cos \alpha ) }{(1 + \cos\alpha) } }{ 1 - \frac{(1 - \cos\theta) }{(1 + \cos\theta )} \times ( \frac{1 - \cos \alpha }{1 + \cos \alpha } )} = \frac{2( \cos \alpha - \cos\theta) }{2( \cos \alpha + \cos\theta) } [/tex]
[tex] = \frac{ \cos \alpha (1 - \cos \beta ) }{ \cos \alpha (1 + \cos \beta ) } (∵ \cos\theta = \cos \alpha \cos \beta ) = { \tan }^{2} \frac{ \beta }{2} [/tex]
Therefore, ur answer is [tex]2) \boxed{ { \tan }^{2} \frac{ \beta }{2} }[/tex]