Question

Sin A tan A/ 1 - cos A = 1+ sec A​

Answer 1

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

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