Question

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

Answer 1

Appropriate Question :

• Prove that : $\longmapsto \:\:\bf \dfrac{ Cos A }{1 - Tan A } - \dfrac{Sin A }{1- Cot A } = (Sin A + Cos A )$

⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━⠀

Given : $\longmapsto \:\:\bf \dfrac{ Cos A }{1 - Tan A } - \dfrac{Sin A }{1- Cot A } = (Sin A + Cos A )$

Exigency To Prove : $\longmapsto \:\:\sf \dfrac{ Cos A }{1 - Tan A } - \dfrac{Sin A }{1- Cot A } = (Sin A + Cos A )$

⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━⠀

$\qquad \quad \bigstar\:\: \:\:\bf \dfrac{ Cos A }{1 - Tan A } - \dfrac{Sin A }{1- Cot A } = (Sin A + Cos A )$

Here ,

• $\longmapsto \:\bf{L.H.S}\:=\:\sf \dfrac{ Cos A }{1 - Tan A } - \dfrac{Sin A }{1- Cot A }$
• $\longmapsto \:\bf{R.H.S}\:=\:\sf Sin A + Cos A$

⠀⠀⠀⠀⠀⠀$\underline {\frak{\star\:Now \: By \: Solving \: \: Given \: L.H.S \::}}$

$\longmapsto \:\bf{L.H.S}\:=\:\sf \dfrac{ Cos A }{1 - Tan A } - \dfrac{Sin A }{1- Cot A }$

$:\implies \:\:\sf \dfrac{ Cos A }{1 - Tan A } - \dfrac{Sin A }{1- Cot A }$

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

• $Tan A = \dfrac{Sin A }{Cos A}$

$:\implies \:\:\sf \dfrac{ Cos A }{1 - \dfrac{Sin A }{CosA} } - \dfrac{Sin A }{1- Cot A }$

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

• $Cot A = \dfrac{Cos A }{Sin A}$

$:\implies \:\:\sf \dfrac{ Cos A }{1 - \dfrac{Sin A }{CosA} } - \dfrac{Sin A }{1- \dfrac{Cos A }{Sin A } }$

$:\implies \:\:\sf \dfrac{ Cos A }{1 - \dfrac{Sin A }{CosA} } - \dfrac{Sin A }{1- \dfrac{Cos A }{Sin A } }$

$:\implies \:\:\sf \dfrac{ Cos A }{ \dfrac{Cos A - Sin A }{CosA} } - \dfrac{Sin A }{ \dfrac{Sin A - Cos A }{Sin A } }$

$:\implies \:\:\sf \dfrac{ Cos A \times Cos A }{ Cos A - Sin A } - \dfrac{Sin A \times Sin A }{ Sin A - Cos A }$

$:\implies \:\:\sf \dfrac{ Cos^2 A }{ Cos A - Sin A } - \dfrac{Sin^2 A }{ Sin A - Cos A }$

$:\implies \:\:\sf \dfrac{ Cos^2 A - Sin^2 A }{ Cos A - Sin A }$

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

• a² - b² = ( a + b ) ( a - b )

$:\implies \:\:\sf \dfrac{ (Cos A - Sin A)(Cos A + Sin A ) }{ Cos A - Sin A }$

$:\implies \:\:\sf \dfrac{ \cancel{(Cos A - Sin A)}(Cos A + Sin A ) }{\cancel {Cos A - Sin A } }$

$:\implies \:\:\bf L.H.S = (Cos A + Sin A )$

⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━⠀

Therefore,

• $\longmapsto \:\bf{L.H.S}\:=\:\sf Sin A + Cos A$
• $\longmapsto \:\bf{R.H.S}\:=\:\sf Sin A + Cos A$

As, we can See that ,

• L.H.S = R.H.S

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

⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━⠀

Answer 2

$\large\underline{\bold{Correct \:Question - }}$

$\sf \: Prove \: that \: \dfrac{cosA}{1 - tanA} + \dfrac{sinA}{1 - cotA} = sinA \: + \: cosA$

$\large\underline{\sf{Solution-}}$

Consider LHS,

$\sf \: \dfrac{cosA}{1 - tanA} + \dfrac{sinA}{1 - cotA}$

We know,

$\boxed{ \bf{ \: tanA = \dfrac{sinA}{cosA} }}$

and

$\boxed{ \bf{ \: cotA = \dfrac{cosA}{sinA} }}$

So,

• LHS can be rewritten as

$\sf \: = \: \dfrac{cosA}{1 - \dfrac{sinA}{cosA} } + \dfrac{sinA}{1 - \dfrac{cosA}{sinA} }$

$\sf \: = \: \dfrac{cosA}{\dfrac{cosA - sinA}{cosA} } + \dfrac{sinA}{\dfrac{sinA - cosA}{sinA} }$

$\sf \: = \: \dfrac{ {cos}^{2}A }{cosA - sinA} + \dfrac{ {sin}^{2} A}{sinA - cosA}$

$\sf \: = \: \dfrac{ {cos}^{2}A }{cosA - sinA} - \dfrac{ {sin}^{2} A}{cosA - sinA}$

$\sf \: = \: \dfrac{ {cos}^{2}A - {sin}^{2}A }{cosA - sinA}$

$\sf \: = \: \dfrac{(cosA + sinA) \cancel{(cosA - sinA)}}{ \cancel{(cosA - sinA)}} \: \: \{ \because{a}^{2} - {b}^{2} = (a + b)(a - b) \}$

$\sf \: = \: cosA + sinA$

$\sf \: = \: RHS$

${\boxed{\boxed{\bf{Hence, Proved}}}}$

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