Question

prove that, cosA/1+tanA+sinA/1-cotA=cosA+sinA ​

Answer 1

Need To Prove :-

• $\begin{gathered} \\ \sf \green{\dfrac{cosA}{1 - tanA} + \dfrac{sinA}{1 - cotA} = cosA + sinA}\end{gathered}$

Formula's :-

• $\begin{gathered} \\ \sf \purple {\: tanx = \frac{sinx}{cosx} }\end{gathered}$
• $\begin{gathered} \\ \sf\purple{\: cotx = \frac{1}{tanx} = \frac{cosx}{sinx}} \end{gathered}$

• $\begin{gathered} \\ \sf\purple{ \: secx = \frac{1}{cosx} }\end{gathered}$

Proof :-

$\sf\pink{:\implies L.H.S= \dfrac{cosA}{1 - tanA} + \dfrac{sinA}{1 - cotA}} \\ \\\\ { \sf:\implies \dfrac{ \cos A}{1 - \dfrac{ \sin A}{ \cos A} \: + \: \dfrac{ \sin A}{1 - \dfrac{ \cos A}{ \sin A} } }} \\ \\ \\ \sf :\implies \dfrac{ \cos A}{ \dfrac{ \cos A - \sin A}{ \cos A} } \: + \: \dfrac{ \sin A}{ \dfrac{ \sin A - \cos A}{ \sin A} } \\ \\\\ \sf: \implies \dfrac{ \cos^{2}A}{ \cos A - \sin A} \: + \: \dfrac{ \sin^{2}A }{ \sin A - \cos A} \\ \\\\ \sf: \implies \dfrac{ \cos^{2} A}{ \cos A - \sin A} \: - \: \dfrac{ \sin^{2}A }{ \cos A - \sin A} \\\\\\ \sf :\implies \dfrac{ \cos^{2}A - \sin^{2}A }{ \cos A - \sin A} \\ \\\\ \sf: \implies \dfrac{( \cos A - \sin A)( \cos A + \sin A)}{ \cos A - \sin A} \\ \\\\ \sf \pink{:\implies \cos A + \sin A \: = \: RHS}$

• Hence,( Proved..!)

Answer 2

Appropriate Question :

⠀⠀⠀⠀⠀▪︎ Prove that : $\sf \leadsto \dfrac{cos A }{1 - tan A } + \dfrac{ sin A }{1 - cot A} = cos A + sin A$

⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━⠀

Given : $\sf \leadsto \dfrac{cos A }{1 - tan A } + \dfrac{ sin A }{1 - cot A} = cos A + sin A$

Exigency To Prove : $\sf \leadsto \dfrac{cos A }{1 - tan A } + \dfrac{ sin A }{1 - cot A} = cos A + sin A$ [ L.H.S = R.H.S ]

⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━⠀

$\qquad \dag\:\:\bigg\lgroup \sf{ \dfrac{cos A }{1 - tan A } + \dfrac{ sin A }{1 - cot A} = cos A + sin A }\bigg\rgroup$

Here ,

$\qquad \leadsto \bf L.H.S \: = \: \sf \dfrac{cos A }{1 - tan A } + \dfrac{ sin A }{1 - cot A}$

$\qquad \leadsto \bf R.H.S \: = \: \sf cos A + sin A$

⠀⠀⠀⠀⠀⠀$\underline {\boldsymbol{\star\:Now \: By \: Solving \: the \: L.H.S\::}}$

$\qquad :\implies \bf L.H.S \: = \: \sf \dfrac{cos A }{1 - tan A } + \dfrac{ sin A }{1 - cot A}$

$\qquad :\implies \sf \dfrac{cos A }{1 - tan A } + \dfrac{ sin A }{1 - cot A}$

$\dag\:\:\sf{ As,\:We\:know\:that\::}$

$\qquad \dag\:\:\bigg\lgroup \sf{ tan \theta = \dfrac{ sin \theta }{ cos \theta } }\bigg\rgroup$

$\qquad :\implies \sf \dfrac{cos A }{1 -tan A } + \dfrac{ sin A }{1 - cot A}$

$\qquad :\implies \sf \dfrac{cos A }{1 - \dfrac{ sin A }{ cos A }} + \dfrac{ sin A }{1 - cot A}$

$\dag\:\:\sf{ As,\:We\:know\:that\::}$

$\qquad \dag\:\:\bigg\lgroup \sf{ cot \theta = \dfrac{ cos \theta }{ sin \theta } }\bigg\rgroup$

$\qquad :\implies \sf \dfrac{cos A }{1 - \dfrac{ sin A }{ cos A }} + \dfrac{ sin A }{1 - cot A}$

$\qquad :\implies \sf \dfrac{cos A }{1 - \dfrac{ sin A }{ cos A }} + \dfrac{ sin A }{1 - \dfrac{ cos A }{ sin A } }$

$\qquad :\implies \sf \dfrac{cos A }{ \dfrac{cos A - sin A }{ cos A }} + \dfrac{ sin A }{ \dfrac{ sin A - cos A }{ sin A } }$

$\qquad :\implies \sf \dfrac{cos^2 A }{ cos A - sin A } + \dfrac{ sin^2 A }{ sin A - cos A }$

$\qquad :\implies \sf \dfrac{cos^2 A - sin^2 A }{ cos A - sin A }$

$\dag\:\:\sf{ As,\:We\:know\:that\::}$

$\qquad \dag\:\:\bigg\lgroup \sf{ Algebraic \:Indentity\:\:= a^2 - b^2 = ( a + b ) ( a - b) }\bigg\rgroup$

$\qquad :\implies \sf \dfrac{cos^2 A - sin^2 A }{ cos A - sin A }$

$\qquad :\implies \sf \dfrac{(cos A - sin A)(cos A + sin A ) }{ cos A - sin A }$

$\qquad :\implies \sf \dfrac{\cancel{(cos A - sin A)}(cos A + sin A ) }{ \cancel {(cos A - sin A)} }$

$\qquad :\implies \sf (cos A + sin A )$

$\qquad \bf L.H.S \:= \sf cos A + sin A$

Therefore,

$\qquad \leadsto \bf L.H.S \: = \: \sf cos A + sin A$

$\qquad \leadsto \bf R.H.S \: = \: \sf cos A + sin A$

⠀⠀⠀⠀⠀As , We can clearly see that L.H.S = R.H.S ;

⠀⠀⠀⠀⠀$\therefore {\underline {\bf{ Hence, \:Proved \:}}}$

⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━⠀