Question

cosa/1+tana-sina/1+cota=cosa-sina​

Answer 1

Answer:

Step-by-step explanation:

LHS =cosA/(1-tanA)+sinA/(1-cotA)

       =cos A/(1 - sin A/cos A) + sin A/(1 - cos A/sin A)

       =cos²A/ (cos A - sin A) + sin²A / (sin A - cos A)

       =cos²A/ (cos A - sin A) - sin²A / (cos A - sin A)

       =(cos ² A - sin ² A) / (cos A - sin A)

       =(cos A - sin A)(cos A + sin A) / (cos A - sin A)

       =cos A + sin A

Answer 2

Answer:

Since

$( \tan(a) = \frac{ \sin(a) }{ \cos(a) })$

&

$\cot(a) = \frac{1}{ \tan(a) }$

$lhs = \frac{ \cos(a) }{1 + \frac{ \sin(a) }{ \cos(a) } } - \frac{ \sin(a) }{1 + \frac{ \cos(a) }{ \sin(a) } }$

$= \frac{ \cos(a) }{ \frac{ \cos(a) + \sin(a) }{ \cos(a) } } - \frac{ \sin(a) }{ \frac{ \sin(a) + \cos(a) }{ \sin(a) } }$

$= \: \frac{ { \cos(a) }^{2} - { \sin(a) }^{2} }{ \sin(a )+ \cos(a) }$

${since \: {a}^{2} - {b}^{2} = (a + b)( a- b)}$

$= \frac{( \cos(a) + \sin(a) )( \cos(a) - \sin(a) )}{( \cos(a) + \sin(a) ) }$

$= \cos(a) - \sin(a)$

$|hence \: proved \: |$

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