prove that
Cos A/1-Tan A – Sin A / 1-Cot A
= Sin A – Cos A

2 thoughts on “prove that <br /> Cos A/1-Tan A – Sin A / 1-Cot A<br /> = Sin A – Cos A”

1. [tex]\large\underline{\bold{Correct \:Question – }}[/tex]

   [tex] \sf \: Prove \: that \: \dfrac{cosA}{1 – tanA} + \dfrac{sinA}{1 – cotA} = sinA \: + \: cosA[/tex]

   [tex]\large\underline{\sf{Solution-}}[/tex]

   Consider LHS,

   [tex] \sf \: \dfrac{cosA}{1 – tanA} + \dfrac{sinA}{1 – cotA}[/tex]

   We know,

   [tex] \boxed{ \bf{ \: tanA = \dfrac{sinA}{cosA} }}[/tex]

   and

   [tex] \boxed{ \bf{ \: cotA = \dfrac{cosA}{sinA} }}[/tex]

   So,

   • LHS can be rewritten as

   [tex] \sf \: = \: \dfrac{cosA}{1 – \dfrac{sinA}{cosA} } + \dfrac{sinA}{1 – \dfrac{cosA}{sinA} }[/tex]

   [tex] \sf \: = \: \dfrac{cosA}{\dfrac{cosA – sinA}{cosA} } + \dfrac{sinA}{\dfrac{sinA – cosA}{sinA} }[/tex]

   [tex] \sf \: = \: \dfrac{ {cos}^{2}A }{cosA – sinA} + \dfrac{ {sin}^{2} A}{sinA – cosA} [/tex]

   [tex] \sf \: = \: \dfrac{ {cos}^{2}A }{cosA – sinA} – \dfrac{ {sin}^{2} A}{cosA – sinA} [/tex]

   [tex] \sf \: = \: \dfrac{ {cos}^{2}A – {sin}^{2}A }{cosA – sinA} [/tex]

   [tex] \sf \: = \: \dfrac{(cosA + sinA) \cancel{(cosA – sinA)}}{ \cancel{(cosA – sinA)}} \: \: \{ \because{a}^{2} – {b}^{2} = (a + b)(a – b) \}[/tex]

   [tex] \sf \: = \: cosA + sinA[/tex]

   [tex] \sf \: = \: RHS[/tex]

   [tex]{\boxed{\boxed{\bf{Hence, Proved}}}}[/tex]

   Additional Information:-

   Relationship between sides and T ratios

   sin θ = Opposite Side/Hypotenuse

   cos θ = Adjacent Side/Hypotenuse

   tan θ = Opposite Side/Adjacent Side

   sec θ = Hypotenuse/Adjacent Side

   cosec θ = Hypotenuse/Opposite Side

   cot θ = Adjacent Side/Opposite Side

   Reciprocal Identities

   cosec θ = 1/sin θ

   sec θ = 1/cos θ

   cot θ = 1/tan θ

   sin θ = 1/cosec θ

   cos θ = 1/sec θ

   tan θ = 1/cot θ

   Co-function Identities

   sin (90°−x) = cos x

   cos (90°−x) = sin x

   tan (90°−x) = cot x

   cot (90°−x) = tan x

   sec (90°−x) = cosec x

   cosec (90°−x) = sec x

   Fundamental Trigonometric Identities

   sin²θ + cos²θ = 1

   sec²θ – tan²θ = 1

   cosec²θ – cot²θ = 1

   Reply

2. Appropriate Question :

   • Prove that : [tex]\longmapsto \:\:\bf \dfrac{ Cos A }{1 – Tan A } – \dfrac{Sin A }{1- Cot A } = (Sin A + Cos A ) \\ \\ \\ [/tex]

   ⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━⠀

   Given : [tex]\longmapsto \:\:\bf \dfrac{ Cos A }{1 – Tan A } – \dfrac{Sin A }{1- Cot A } = (Sin A + Cos A ) \\ \\ \\ [/tex]

   Exigency To Prove : [tex]\longmapsto \:\:\sf \dfrac{ Cos A }{1 – Tan A } – \dfrac{Sin A }{1- Cot A } = (Sin A + Cos A ) \\ \\ \\ [/tex]

   ⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━⠀

   [tex]\qquad \quad \bigstar\:\: \:\:\bf \dfrac{ Cos A }{1 – Tan A } – \dfrac{Sin A }{1- Cot A } = (Sin A + Cos A ) \\ \\ \\ [/tex]

   Here ,

   • [tex]\longmapsto \:\bf{L.H.S}\:=\:\sf \dfrac{ Cos A }{1 – Tan A } – \dfrac{Sin A }{1- Cot A } \\ \\ \\ [/tex]
   • [tex]\longmapsto \:\bf{R.H.S}\:=\:\sf Sin A + Cos A \\ \\ \\ [/tex]

   ⠀⠀⠀⠀⠀⠀[tex]\underline {\frak{\star\:Now \: By \: Solving \: \: Given \: L.H.S \::}}\\[/tex]

   [tex]\longmapsto \:\bf{L.H.S}\:=\:\sf \dfrac{ Cos A }{1 – Tan A } – \dfrac{Sin A }{1- Cot A } \\ \\ \\ [/tex]

   [tex]:\implies \:\:\sf \dfrac{ Cos A }{1 – Tan A } – \dfrac{Sin A }{1- Cot A } \\ \\ \\ [/tex]

   [tex]\dag\:\it{As,\:We\:know\:that\::}\\\\[/tex]

   • [tex] Tan A = \dfrac{Sin A }{Cos A}\\\\[/tex]

   [tex]:\implies \:\:\sf \dfrac{ Cos A }{1 – \dfrac{Sin A }{CosA} } – \dfrac{Sin A }{1- Cot A } \\ \\ \\ [/tex]

   [tex]\dag\:\it{As,\:We\:know\:that\::}\\\\[/tex]

   • [tex] Cot A = \dfrac{Cos A }{Sin A}\\\\[/tex]

   [tex]:\implies \:\:\sf \dfrac{ Cos A }{1 – \dfrac{Sin A }{CosA} } – \dfrac{Sin A }{1- \dfrac{Cos A }{Sin A } } \\ \\ \\ [/tex]

   [tex]:\implies \:\:\sf \dfrac{ Cos A }{1 – \dfrac{Sin A }{CosA} } – \dfrac{Sin A }{1- \dfrac{Cos A }{Sin A } } \\ \\ \\ [/tex]

   [tex]:\implies \:\:\sf \dfrac{ Cos A }{ \dfrac{Cos A – Sin A }{CosA} } – \dfrac{Sin A }{ \dfrac{Sin A – Cos A }{Sin A } } \\ \\ \\ [/tex]

   [tex]:\implies \:\:\sf \dfrac{ Cos A \times Cos A }{ Cos A – Sin A } – \dfrac{Sin A \times Sin A }{ Sin A – Cos A } \\ \\ \\ [/tex]

   [tex]:\implies \:\:\sf \dfrac{ Cos^2 A }{ Cos A – Sin A } – \dfrac{Sin^2 A }{ Sin A – Cos A } \\ \\ \\ [/tex]

   [tex]:\implies \:\:\sf \dfrac{ Cos^2 A – Sin^2 A }{ Cos A – Sin A } \\ \\ \\ [/tex]

   [tex]\dag\:\it{As,\:We\:know\:that\::}\\\\[/tex]

   • a² – b² = ( a + b ) ( a – b )

   [tex]:\implies \:\:\sf \dfrac{ (Cos A – Sin A)(Cos A + Sin A ) }{ Cos A – Sin A } \\ \\ \\ [/tex]

   [tex]:\implies \:\:\sf \dfrac{ \cancel{(Cos A – Sin A)}(Cos A + Sin A ) }{\cancel {Cos A – Sin A } } \\ \\ \\ [/tex]

   [tex]:\implies \:\:\bf L.H.S = (Cos A + Sin A ) \\ \\ \\ [/tex]

   ⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━⠀

   Therefore,

   • [tex]\longmapsto \:\bf{L.H.S}\:=\:\sf Sin A + Cos A \\ \\ \\ [/tex]
   • [tex]\longmapsto \:\bf{R.H.S}\:=\:\sf Sin A + Cos A \\ \\ \\ [/tex]

   As, we can See that ,

   • L.H.S = R.H.S

   ⠀⠀⠀⠀⠀[tex]\therefore {\underline {\bf{ Hence, \:Proved ! \:}}}\\\\\\[/tex]

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