Question

$\sf{If\:tan\:\dfrac{\theta}{2} = \sqrt{\dfrac{a-b}{a+b} } \:tan\dfrac{\varnothing}{2} \:,\:prove\:that \\\ cos\:\theta=\dfrac{a\:cos\:\varnothing+b}{a+b\:cos\:\varnothing}}$

Answer 1

AnswEr :

Given that,

$\sf \: tan( \dfrac{ \theta}{2} ) = \sqrt{ \dfrac{a - b}{a + b} } tan( \dfrac{ \phi}{2} )$

To Prove :

$\sf \: cos \: \theta = \dfrac{acos \: \phi + b}{a + bcos \: \phi}$

We Know that,

$\sf \: cos2x = \dfrac{1 - tan {}^{2}x }{1 + tan {}^{2}x }$

Similarly,

$\longrightarrow \sf cos \ \theta = \dfrac{1 - tan^2 \frac{\theta}{2}}{1 + tan^2 \frac{\theta}{2}}$

Then,

$\sf \: cos \ \theta = \dfrac{1 - ( \sqrt{ \dfrac{a - b}{a + b} } \: tan \frac{ \phi}{2} ) {}^{2} }{1 + ( \sqrt{ \dfrac{a - b}{a + b} } \: tan \frac{ \phi}{2}) {}^{2} } \\ \\ \\ \longrightarrow \: \sf \: cos \: \theta = \frac{1 - \bigg( \dfrac{a - b}{a + b} \times \dfrac{ {sin}^{2} \frac{ \phi}{2} }{ {cos}^{2} \frac{ \phi}{2} } \bigg)}{1 + \bigg( \dfrac{a - b}{a + b} \times \dfrac{ {sin}^{2} \frac{ \phi}{2} }{ {cos}^{2} \frac{ \phi}{2} } \bigg) } \\ \\ \\ \longrightarrow \: \sf \: cos \: \theta = \dfrac{(a + b) {cos}^{2} \frac{ \phi}{2} - (a - b) {sin}^{2} \frac{ \phi}{2} }{(a + b) {cos}^{2} \frac{ \phi}{2} + (a - b) {sin}^{2} \frac{ \phi}{2} } \\ \\ \\ \longrightarrow \: \sf \: cos \: \theta = \dfrac{a(cos {}^{2} \frac{ \phi}{2} - {sin}^{2} \frac{ \phi}{2}) + b(cos {}^{2} \frac{ \phi}{2} + {sin}^{2} \frac{ \phi}{2} ) }{a(cos {}^{2} \frac{ \phi}{2} + {sin}^{2} \frac{ \phi}{2} ) + b(cos {}^{2} \frac{ \phi}{2} - {sin}^{2} \frac{ \phi}{2} ) } \\ \\ \\ \longrightarrow \: \sf \: cos \: \theta = \frac{acos \: \phi + b}{a + bcos \: \phi}$

Hence ProveD

Identities Used :

• cos²∅ - sin²∅ = cos2∅

• sin²∅ + cos²∅ = 1

Answer 2

$\large \underline{ \blue{ \boxed{ \bf \green{Required \: Answer}}}}$

$\sf{If\:tan\:\dfrac{\theta}{2} = \sqrt{\dfrac{a-b}{a+b} } \:tan\dfrac{\varnothing}{2} \:,\:prove\:that \\\ cos\:\theta=\dfrac{a\:cos\:\varnothing+b}{a+b\:cos\:\varnothing}}$