Question

NCERT 2020
CLASS 12.
(TRIGONOMETRY)

Solve if you know..

Irrelevant answers will be directly Reported✏️✒️​

Answer 1

Question:-

• If $\sf \tan( \frac{x}{2} ) = \sqrt{ \frac{1 - e}{1 + e} } \times \tan( \frac{ \alpha }{2} )$then find the value of $\sf \cos \alpha$

Formula to be used here:-

• $\sf \tan( \frac{x}{2} ) = \sqrt{ \frac{1 - \cos(x) }{1 + \cos(x) } }$

Solution:-

• Check out the attachment. Solution is provided there.

$\sf \tan( \frac{x}{2} ) = \sqrt{ \frac{1 - e}{1 + e} } \times \tan( \frac{ \alpha }{2} )$

$\sf \implies \sqrt{ \frac{1 - \cos(x) }{1 + \cos(x) } } = \sqrt{ \frac{1 - e}{1 + e} } \times \sqrt{ \frac{1 - \cos( \alpha ) }{1 + \cos( \alpha ) } }$

Squaring both side, we get,

$\sf \frac{1 - \cos(x) }{1 + \cos(x) } = \frac{(1 - e)(1 - \cos( \alpha )) }{(1 + e)(1 + \cos( \alpha )) }$

On cross multiplying and simplifying, we get,

$\sf 2 \cos( \alpha ) + 2e = 2 \cos(x) + 2e \cos(x) \cos(\alpha )$

$\sf \implies 2 \cos( \alpha ) - 2e \cos( \alpha ) \cos(x) =2 \cos(x) - 2e$

$\sf \implies \cos( \alpha ) - e \cos( \alpha ) \cos(x) = \cos(x) - e$

$\sf \implies \cos( \alpha ) (1 - e \cos(x) )= \cos(x) - e$

$\sf \implies \cos( \alpha )= \frac{\cos(x) - e}{1 - e \cos(x) }$

Answer:-

$\boxed{ \sf \cos( \alpha )= \frac{\cos(x) - e}{1 - e \cos(x) }}$

Answer 2

$\bold{QuestioN:}$

$\bold{if\:tan\:\frac{x}{2}=\sqrt{\frac{1-e}{1+e} }\:tan \:\frac{\alpha }{2},\:then\:what\:is\:cos\alpha ?? }$

$\huge\fcolorbox{black}{lime}{AnsweR:}$  ^_^

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$\bold{(tan\:(\frac{x}{2} ))^2 = (\sqrt{\frac{1-e}{1+e} } )^2\:tan\:(\frac{\alpha}{2}) }$,

$\bold{tan^2\:\frac{x}{2}=\frac{1-e}{1+e}\:tan^2\:\frac{\alpha }{2} }$,

$\bold{tan^2\:\frac{\alpha }{2} }$  =  $\small\boxed{\frac{(1+e)\:tan^2\:(\frac{x}{2} )}{(1-e)} }$,

$\bold{cos\:\alpha =}$ $\boxed{\frac{1-tan^2\:\frac{\alpha }{2} }{1+tan^2\:\frac{\alpha }{2} } }$,

$\large\boxed{=\frac{\frac{1-(1+e)\:tan^2\:\frac{x }{2} }{(1-e)} }{1+\frac{(1+e)}{(1-e) }tan^2\:\frac{x }{2} } }$,

$\bold{cos\:\alpha =}$ $\boxed{\frac{(1-e)-(1+e)\:tan^2\:\frac{x}{2} }{(1-e)+(1+e)\:tan^2\:\frac{x}{2}} }$

$\bold{cos\:\alpha =}$   $\large\boxed{\frac{(1-tan^2\frac{x}{2}-e(1+tan^2\frac{x}{2} ) )}{(1+tan^2\frac{x}{2}-e(1-tan^2\frac{x}{2} ) )} }$

$\bold{cos\:\alpha =}$  $\large\boxed{\frac{(\frac{1-tan^2\frac{x}{2} }{1+tan^2\frac{x}{2} } )-e}{1-e(\frac{1-tan^2\frac{x}{2} }{1+tan^2\frac{x}{2} } )} }$

$\bold{So,\:cos\alpha = }$  $\large\boxed{\frac{cos\:x-e}{1-ecos\:x}}$.

$\large\fcolorbox{black}{lime}{Hope\:my\:answer\:is\:short\:but\:correct\:and\:perfect}$

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