3 years ago

HELP PLEASE.................

Mathematics

1 answer:

Kryger [21]3 years ago

6 0

Zero

There is no solution for these equations as there is no even one point of intersection between them

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Sin A tan A/ 1 - cos A = 1+ sec A

aalyn [17]

$\text{L.H.S}\\\\=\dfrac{\sin A \tan A}{1- \cos A}\\\\\\=\dfrac{\sin A \cdot \tfrac{\sin A}{\cos A} }{ 1- \cos A}~~~~~~~~~~~~~~~~~~~~~~~~~~~~~;\left[\tan A = \dfrac{\sin A }{\cos A}\right]\\\\\\=\dfrac{\tfrac{\sin^2 A}{ \cos A}}{1- \cos A}\\\\\\=\dfrac{\sin^2 A}{\cos A(1-\cos A)}\\\\\\=\dfrac{1-\cos^2 A}{\cos A( 1- \cos A)}~~~~~~~~~~~~~~~~~~~~~~~~;[\sin^2 A + \cos^2 A = 1]\\\\\\$

$=\dfrac{(1+ \cos A)(1 - \cos A)}{\cos A(1 - \cos A)}\\\\\\=\dfrac{1+ \cos A}{ \cos A}\\\\\\=\dfrac{1}{\cos A} + \dfrac{\cos A}{ \cos A}\\\\\\=\sec A + 1 \\\\\\=1+ \sec A\\\\\\=\text{R.H.S}\\\\\text{Proved.}$