Question

prove that
1+cosa/sina = sina/1-cosa​

Answer 1

Answer:

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Answer 2

Answer:

multiplying and dividing by

$1 - \cos( \alpha )$

then,

$lhs \\ \frac{1 + \cos( \alpha ) }{ \sin( \alpha ) } \\ = \frac{1 + \cos( \alpha ) }{ \sin( \alpha ) } \times \frac{1 - \cos( \alpha ) }{1 - \cos( \alpha ) } \\ = \frac{ {1}^{2} - { \cos( \alpha ) }^{2} }{ \sin \alpha (1 - \cos\alpha ) } \\ = \frac{1 - { \cos \alpha }^{2} }{ \sin \alpha(1 - \cos\alpha ) } \\ = \frac{ { \sin( \alpha ) }^{2} }{ \sin\alpha (1 - \cos\alpha ) } \\ = \frac{ \sin( \alpha ) }{1 - \cos( \alpha ) } = rhs$

proved...

Step-by-step explanation:

formula used : ( a + b)( a -b)= a² -b²

and

sin²∅ + cos²∅ = 1

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