Question

$\displaystyle{If \ ;}\\\\\displaystyle{\tan\frac{\theta}{2}=\sqrt{\dfrac{1-e}{1+e}}\tan\frac{\phi}{2}}\\\\\\\displaystyle{Prove \ that \ :}\\\\\\\displaystyle{\cos \phi=\dfrac{\cos\theta-e}{1-e \ \cos\theta}}$

Answer 1

Stuart Little

Here is your answer:-

Refer the attachment

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Answer 2

SOLUTION:-

To prove:

$\cos \theta = \frac{cos \theta - e}{1 - e \: cos \theta}$

Proof:

L.H.S

$cos \theta = \frac{1 - tan2 \frac{ \theta \: }{2} }{1 + tan \: 2 \frac{ \theta}{2} } \: \: \: \: [As \: cos \: 2 \theta = \frac{1 - tan {}^{2} \theta}{1 + tan \: {}^{2} \theta } ]..........(1) \\ \\ ∵ tan( \frac{ \theta}{2} ) = \sqrt{ \frac{(1 - e)}{(1 + e)} } tan[ \frac{ \theta}{2} ] \: \: \: \: \: (given)$

Squaring both sides and rearranging,

$</u></strong><strong><u>∴</u></strong><strong><u>tan {}^{2} ( \frac{ \theta}{2} ) = \frac{( 1 + e)}{(1 - e)} tan \: {}^{2} \frac{ \theta}{2} ............(2) \\ putting \: {tan}^{2} ( \frac{ \theta}{2} ) \: from \: eq.2 \: in \: eq.1 \\ </u></strong><strong><u>∴</u></strong><strong><u> \frac{1 - tan \: {}^{2} ( \frac{ \theta}{2} )}{1 + tan \: {}^{2}( \frac{ \theta}{2} )} = \frac{1 - ( \frac{(1 + e)}{( 1 - e)} tan \: {}^{2} ( \frac{ \theta}{2} ))}{1 + ( \frac{(1 + e)}{(1 - e)} tan \: {}^{2} ( \frac{ \theta}{2})) } \\ \\ = > \frac{1 - e - tan \: {}^{2}( \frac{ \theta}{2} ) - e \: tan \: {}^{2} ( \frac{ \theta}{2} )}{1 - e + tan \: {}^{2} ( \frac{ \theta}{2}) + e \: tan \: {}^{2}( \frac{ \theta}{2} ) } \\ \\ = > \frac{(1 - tan \: {}^{2} ( \frac{ \theta}{2})) - e(1 + tan \: {}^{2}( \frac{ \theta}{2} )) }{(1 + tan \: {}^{2}( \frac{ \theta}{2})) - e(1 - tan \: {}^{2} ( \frac{ \theta}{2} ))}$

Dividing numerator & denominator by (1+tan ²theta2).

$= > \frac{( \frac{(1 - tan \: {}^{2} \frac{ \theta}{2}) }{(1 + tan \: {}^{2} \frac{ \theta}{2}) }) - e( \frac{(1 + tan \: {}^{2} \frac{ \theta}{2} )}{(1 + tan \: {}^{2} \frac{ \theta}{2}) } ) }{( \frac{(1 + tan \: {}^{2} \frac{ \theta}{2} ) }{(1 + tan \: {}^{2} \frac{ \theta}{2} ) } - e( \frac{(1 - tan \: {}^{2} \frac{ \theta}{2} )}{(1 + tan \: {}^{2} \frac{ \theta}{2}) } ) } \\ \\ = > ( \frac{cos \theta - e}{1 - e \: cos \theta} ) \: \: \: \: \: \: [cos \: 2 \theta = \frac{1 - tan \: {}^{2} \theta}{1 + tan {}^{2} \theta} ]$

R.H.S

Hence, proved

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