Question

please answer this quality answer will be marked as brainliest
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explain briefly​

Answer 1

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

$\green{\tt{\therefore{tan \frac{ \theta + \alpha }{2} tan \frac{ \theta - \alpha }{2} = {tan}^{2} \: \frac{ \beta }{2}}}}$

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

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

• According to given question :

$\bold{As \: we \: know \: that} \\ \tt: \implies\tan \frac{ \theta + \alpha }{2} tan \frac{ \theta - \alpha }{2} \\ \\ \tt: \implies \frac{sin \: \bigg (\frac{ \theta + \alpha }{2} \bigg)}{cos \bigg( \frac{ \theta + \alpha }{2} \bigg)} \times \frac{sin \bigg(\frac{ \theta - \alpha }{2} \: \bigg)}{cos \: \bigg(\frac{ \theta - \alpha }{2} \bigg)} \\ \\ \text{Multiplying \: both \: in \: numerator \: and \: denominator \:by \: 2} \\ \\ \tt: \implies \frac{2 \: sin \: \bigg (\frac{ \theta + \alpha }{2} \bigg)}{2 \: cos \bigg( \frac{ \theta + \alpha }{2} \bigg)} \frac{sin \bigg(\frac{ \theta - \alpha }{2} \: \bigg)}{cos \: \bigg(\frac{ \theta - \alpha }{2} \bigg)}$

$\tt \circ \:2 \: sin \: A\: sin \: B = cos (A - B) - cos(A + B) \\ \\ \tt \circ \:2 \: cos\: A \: cos\: B = cos (A + B) + cos(A - B) \\ \\ \tt: \implies \frac{cos \bigg( \frac{ \theta + \alpha }{2} - \bigg( \frac{ \theta - \alpha }{2} \bigg) \bigg) - cos \bigg( \frac{ \theta + \alpha }{2} + \frac{ \theta - \alpha }{2} \bigg ) }{cos \bigg( \frac{ \theta + \alpha }{2} + \frac{ \theta - \alpha }{2} \bigg) + cos \bigg( \frac{ \theta + \alpha }{2} - \bigg( \frac{ \theta - \alpha }{2} \bigg)\bigg )} \\ \\ \tt: \implies \frac{cos \bigg( \frac{ \theta + \alpha - \theta + \alpha }{2} \bigg) - cos \bigg( \frac{ \theta + \alpha }{2} + \frac{ \theta - \alpha }{2} \bigg ) }{cos \bigg( \frac{ \theta + \alpha }{2} + \frac{ \theta - \alpha }{2} \bigg) + cos \bigg( \frac{ \theta + \alpha - \theta + \alpha }{2} \bigg )}$

$\tt: \implies \frac{cos \: \alpha - cos \: \theta}{cos \: \theta + cos \: \alpha } \\ \\ \tt: \implies \frac{cos \: \alpha - cos \: \alpha \:cos \: \beta }{cos \: \alpha \: cos \: \beta + cos \: \alpha } \\ \\ \tt: \implies \frac{cos \alpha (1 - cos \: \beta) }{cos \: \alpha (cos \: \beta + 1)} \\ \\ \tt: \implies \frac{1 - cos \: \beta }{cos \: \beta + 1} \\ \\ \tt \circ \: cos \: 2 x = 1 - 2{sin}^{2} \:x \\ \\ \tt \circ \: 2x = \beta \implies x = \frac{ \beta }{2} \\ \\ \tt \circ \:cos \: \beta = 1 - 2 {sin}^{2} \: \frac{ \beta }{2} \\ \\ \tt: \implies \frac{1 - \bigg(1 - 2 {sin}^{2} \: \frac{ \beta }{2} \bigg) }{cos \: \beta + 1} \\ \\ \tt: \implies \frac{2 {sin}^{2} \: \frac{ \beta }{2} }{cos \: \beta + 1}$

$\tt \circ \: cos \: 2 x = 2 cos^{2} \: x - 1 \\ \\ \tt \circ \: cos \: \beta = 2 {cos}^{2} \: \frac{ \beta }{2} - 1 \\ \\ \tt: \implies \frac{2sin^{2} \: \frac{ \beta }{2} }{2cos^{2} \: \frac{ \beta }{2} } \\ \\ \tt: \implies \frac{sin^{2} }{cos^{2} } \bigg (\frac{ \beta }{2} \bigg) \\ \\ \green{\tt : \implies {tan}^{2} \: \frac{ \beta }{2} } \\ \\ \green{\tt \therefore tan \frac{ \theta + \alpha }{2} tan \frac{ \theta - \alpha }{2} = {tan}^{2} \: \frac{ \beta }{2} }$

Answer 2

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

$\green{\tt{\therefore{tan \frac{ \theta + \alpha }{2} tan \frac{ \theta - \alpha }{2} = {tan}^{2} \: \frac{ \beta }{2}}}}$

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

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

• According to given question :

$\bold{As \: we \: know \: that} \\ \tt: \implies\tan \frac{ \theta + \alpha }{2} tan \frac{ \theta - \alpha }{2} \\ \\ \tt: \implies \frac{sin \: \bigg (\frac{ \theta + \alpha }{2} \bigg)}{cos \bigg( \frac{ \theta + \alpha }{2} \bigg)} \times \frac{sin \bigg(\frac{ \theta - \alpha }{2} \: \bigg)}{cos \: \bigg(\frac{ \theta - \alpha }{2} \bigg)} \\ \\ \text{Multiplying \: both \: in \: numerator \: and \: denominator \:by \: 2} \\ \\ \tt: \implies \frac{2 \: sin \: \bigg (\frac{ \theta + \alpha }{2} \bigg)}{2 \: cos \bigg( \frac{ \theta + \alpha }{2} \bigg)} \frac{sin \bigg(\frac{ \theta - \alpha }{2} \: \bigg)}{cos \: \bigg(\frac{ \theta - \alpha }{2} \bigg)}$

$\tt \circ \:2 \: sin \: A\: sin \: B = cos (A - B) - cos(A + B) \\ \\ \tt \circ \:2 \: cos\: A \: cos\: B = cos (A + B) + cos(A - B) \\ \\ \tt: \implies \frac{cos \bigg( \frac{ \theta + \alpha }{2} - \bigg( \frac{ \theta - \alpha }{2} \bigg) \bigg) - cos \bigg( \frac{ \theta + \alpha }{2} + \frac{ \theta - \alpha }{2} \bigg ) }{cos \bigg( \frac{ \theta + \alpha }{2} + \frac{ \theta - \alpha }{2} \bigg) + cos \bigg( \frac{ \theta + \alpha }{2} - \bigg( \frac{ \theta - \alpha }{2} \bigg)\bigg )} \\ \\ \tt: \implies \frac{cos \bigg( \frac{ \theta + \alpha - \theta + \alpha }{2} \bigg) - cos \bigg( \frac{ \theta + \alpha }{2} + \frac{ \theta - \alpha }{2} \bigg ) }{cos \bigg( \frac{ \theta + \alpha }{2} + \frac{ \theta - \alpha }{2} \bigg) + cos \bigg( \frac{ \theta + \alpha - \theta + \alpha }{2} \bigg )}$

$\tt: \implies \frac{cos \: \alpha - cos \: \theta}{cos \: \theta + cos \: \alpha } \\ \\ \tt: \implies \frac{cos \: \alpha - cos \: \alpha \:cos \: \beta }{cos \: \alpha \: cos \: \beta + cos \: \alpha } \\ \\ \tt: \implies \frac{cos \alpha (1 - cos \: \beta) }{cos \: \alpha (cos \: \beta + 1)} \\ \\ \tt: \implies \frac{1 - cos \: \beta }{cos \: \beta + 1} \\ \\ \tt \circ \: cos \: 2 x = 1 - 2{sin}^{2} \:x \\ \\ \tt \circ \: 2x = \beta \implies x = \frac{ \beta }{2} \\ \\ \tt \circ \:cos \: \beta = 1 - 2 {sin}^{2} \: \frac{ \beta }{2} \\ \\ \tt: \implies \frac{1 - \bigg(1 - 2 {sin}^{2} \: \frac{ \beta }{2} \bigg) }{cos \: \beta + 1} \\ \\ \tt: \implies \frac{2 {sin}^{2} \: \frac{ \beta }{2} }{cos \: \beta + 1}$

$\tt \circ \: cos \: 2 x = 2 cos^{2} \: x - 1 \\ \\ \tt \circ \: cos \: \beta = 2 {cos}^{2} \: \frac{ \beta }{2} - 1 \\ \\ \tt: \implies \frac{2sin^{2} \: \frac{ \beta }{2} }{2cos^{2} \: \frac{ \beta }{2} } \\ \\ \tt: \implies \frac{sin^{2} }{cos^{2} } \bigg (\frac{ \beta }{2} \bigg) \\ \\ \green{\tt : \implies {tan}^{2} \: \frac{ \beta }{2} } \\ \\ \green{\tt \therefore tan \frac{ \theta + \alpha }{2} tan \frac{ \theta - \alpha }{2} = {tan}^{2} \: \frac{ \beta }{2} }$