Question

prove that ..
cotA - cosA ÷ cot A + cosA = cosecA - 1 ÷cosecA + 1

​

Answer 1

Solution :-

$\sf \dfrac{cotA-cosA}{cotA+cosA}= \dfrac{cosecA-1}{cosecA+1}$

L.H.S :-

$\sf we \ know \ that \ cosA= \dfrac{cotA}{cosecA}\\ \\ \\ \longmapsto \sf \dfrac{cotA-\dfrac{cotA}{cosec A}}{cotA+\dfrac{cotA}{cosecA}}\\ \\ \\ \longmapsto\sf \dfrac{\dfrac{cotA\ cosecA -cotA}{cosecA}}{\dfrac{cosecA\ cotA + cotA}{cosecA}}\\ \\ \\ \longmapsto\sf \dfrac{cosecA cotA-cotA}{\cancel{cosecA}}\times \dfrac{\cancel{cosecA}}{cosecA\ cotA+cotA}\\ \\ \\ \longmapsto\sf \dfrac{cosecAcotA-cotA}{cosecAcotA+cotA}\\ \\ \\ \longmapsto\sf \dfrac{\cancel{cotA}(cosecA-1)}{\cancel{cotA}(cosecA+1)}\\ \\ \\ \longmapsto\sf \dfrac{cosecA-1}{cosecA+1}$

$\sf \dfrac{cosecA-1}{cosecA+1}= \dfrac{cosecA-1}{cosecA+1}\\ \\ \sf \ \ L.H.S= R.H.S \ \ (hence \ proved \ !)$

Answer 2

TO PROVE :

$\longrightarrow\rm{\dfrac{cotA - cosA }{cotA + cosA} = \dfrac{cosecA - 1}{cosecA + 1} }$

PROOF :

$\rm L.H.S. = \dfrac{\cot A - \cos A}{\cot A + \cos A} \\\\ \qquad\underline{\underline{\rm{Identity : \cot A = \dfrac{\cos A}{\sin A} }}} \\\\ \longrightarrow\rm \dfrac{\dfrac{\cos A}{\sin A} - \cos A}{\dfrac{\cos A}{\sin A} + \cos A} \\\\ \longrightarrow\rm \dfrac{\dfrac{\cos A - \sin A \cos A}{\sin A}}{\dfrac{\cos A + \sin A \cos A }{\sin A} } \\\\ \longrightarrow\rm \dfrac{\cos A - \sin A \cos A}{\cancel{\sin A}} \times \dfrac{\cancel{\sin A}}{\cos A + \sin A \cos A} \\\\ \longrightarrow\rm \dfrac{ \cancel{\cos A} (1 - \sin A)}{ \cancel{\cos A} (1 + \sin A)} \\\\ \longrightarrow\rm \dfrac{1 - \sin A}{1 + \sin A} \\\\ \qquad\underline{\underline{\rm{Identity : \sin A = \dfrac{1}{\cosec A} }}} \\\\ \longrightarrow\rm \dfrac{1 - \dfrac{1}{\cosec A} }{1 + \dfrac{1}{\cosec A} } \\\\ \longrightarrow\rm \dfrac{\dfrac{\cosec A - 1 }{\cosec A} }{\dfrac{\cosec A + 1}{\cosec A} } \\\\ \longrightarrow\rm \dfrac{\cosec A - 1}{\cancel{\cosec A}} \times \dfrac{\cancel{\cosec A}}{\cosec A + 1} \\\\ \longrightarrow\underline{\underline{\boxed{\rm{ \dfrac{\cosec A - 1}{\cosec A + 1} }}}} \\\\ \longrightarrow\rm R.H.S. \qquad[Proved]$

$\therefore\underline{\textsf{\textbf{L.H.S. = R.H.S. \quad [Proved] }}}$