Math, asked by nimisha8496, 2 months ago

COS A
Prove That
sin2 A
+
= sin A+cos A
1-tan A sin A-COS A​

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Answers

Answered by Intelligentcat
19

Answer:

\dashrightarrow\:\:\bf  \dfrac{Cos \: A}{1 - Tan \: A} + \dfrac{sin^{2} \: A}{Sin \: A - Cos \: A} = (Cos \: A + Sin \: A) \\ \\

Formulae need to know :

  • {\boxed{\sf{TanA = \dfrac{SinA}{CosA}}}}\\ \\

  • {\boxed{\sf{1 - Sin^{2}A = Cos^{2}A}}}\\ \\

Solution :

\dashrightarrow\:\:\sf  \dfrac{Cos \: A}{1 - Tan \: A} + \dfrac{sin^{2} \: A}{Sin \: A - Cos \: A} \\ \\

\dashrightarrow\:\:\sf  \dfrac{Cos \: A}{1 - \dfrac{Sin \: A}{Cos \: A}} + \dfrac{sin^{2} \: A}{Sin \: A - Cos \: A} \\ \\

\dashrightarrow\:\:\sf  \dfrac{Cos \: A}{\dfrac{Cos \: A - Sin \: A}{Cos \: A}} + \dfrac{sin^{2} \: A}{Sin \: A - Cos \: A}\\ \\

\dashrightarrow\:\:\sf  \dfrac{Cos^{2} \: A}{Cos \: A - Sin \: A} + \dfrac{sin^{2} \: A}{Sin \: A - Cos \: A}\\ \\

\dashrightarrow\:\:\sf  \dfrac{Cos^{2} \: A}{Cos \: A - Sin \: A} - \dfrac{sin^{2} \: A}{Cos \: A - Sin \: A}\\ \\

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

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

 \sf \longrightarrow  \:  \dfrac{(Cos \: A + Sin \: A) {\cancel(Cos \: A - Sin \: A)}^{ \:  \: 1} }{ {\cancel(Cos \: A - Sin \: A)}^{ \: \: } } \\ \\

\dashrightarrow\:\:\sf  \dfrac{(Cos \: A + Sin \: A) (Cos \: A - Sin \: A)}{Cos \: A - Sin \: A} = (Cos \: A + Sin \: A) \\ \\

L.H.S = R.H.S

Hence,

Proved !!

{\boxed{\bf{Verified !}}} \\

Answered by BrainlyRish
13

Given : \longmapsto \:\:\bf \dfrac{Cos \: A}{1 - Tan \: A} + \dfrac{Sin^{2} \: A}{Sin \: A - Cos \: A} = (Cos \: A + Sin \: A) \\ \\\\

Exigency To Prove : \longmapsto \:\:\sf \dfrac{Cos \: A}{1 - Tan \: A} + \dfrac{Sin^{2} \: A}{Sin \: A - Cos \: A} = (Cos \: A + Sin \: A) \\ \\ \\

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\dag\:\: \:\:\bf \dfrac{Cos \: A}{1 - Tan \: A} + \dfrac{sin^{2} \: A}{Sin \: A - Cos \: A} = (Cos \: A + Sin \: A) \\ \\\\

Here ,

  • \bf{L.H.S} \:\:\sf \dfrac{Cos \: A}{1 - Tan \: A} + \dfrac{Sin^{2} \: A}{Sin \: A - Cos \: A}  \\ \\
  • \bf{R.H.S} \:\:\sf Cos \: A + Sin \: A  \\ \\\\

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

\bf{L.H.S} \:=\:\sf \dfrac{Cos \: A}{1 - Tan \: A} + \dfrac{Sin^{2} \: A}{Sin \: A - Cos \: A}  \\ \\\\

\qquad:\implies \:=\:\sf \dfrac{Cos \: A}{1 - Tan \: A} + \dfrac{Sin^{2} \: A}{Sin \: A - Cos \: A}  \\ \\\\

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

  •  tan A = \dfrac{Sin A }{Cos A}\\

\qquad:\implies \:\:\sf \dfrac{Cos \: A}{1 - \purple {Tan \: A}} + \dfrac{Sin^{2} \: A}{Sin \: A - Cos \: A}  \\ \\\\

\qquad:\implies \:\:\sf \dfrac{Cos \: A}{1 - \purple {\dfrac{SinA}{Cos \: A}}} + \dfrac{Sin^{2} \: A}{Sin \: A - Cos \: A}  \\ \\\\

\qquad:\implies \:\:\sf \dfrac{Cos \: A}{1 - \dfrac{SinA}{Cos \: A}} + \dfrac{Sin^{2} \: A}{Sin \: A - Cos \: A}  \\ \\\\

\qquad:\implies \:\:\sf \dfrac{Cos \: A}{ \dfrac{CosA-SinA}{Cos \: A}} + \dfrac{Sin^{2} \: A}{Sin \: A - Cos \: A}  \\ \\ \\

\qquad:\implies \:\:\sf \dfrac{Cos \: A\times Cos A}{ CosA-SinA} + \dfrac{Sin^{2} \: A}{Sin \: A - Cos \: A}  \\ \\ \\

\qquad:\implies \:\:\sf \dfrac{Cos ^2\: A}{ CosA-SinA} + \dfrac{Sin^{2} \: A}{Sin \: A - Cos \: A}  \\ \\\\

\qquad:\implies \:\:\sf \dfrac{Cos ^2\: A}{ CosA-SinA} + \dfrac{Sin^{2} \: A}{Sin \: A - Cos \: A}  \\ \\ \\

\qquad:\implies \:\:\sf \dfrac{Cos ^2\: A}{ CosA-SinA} - \dfrac{Sin^{2} \: A}{Sin \: A - Cos \: A}  \\ \\ \\

\qquad:\implies \:\:\sf \dfrac{Cos ^2\: A- Sin^2 A }{ CosA-SinA}   \\ \\\\

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

  • a² - b² = (a + b) (a - b)

\qquad:\implies \:\:\sf \dfrac{(Cos\: A+ Sin A)(CosA - Sin A ) }{ (CosA-SinA)}   \\\\ \\

\qquad:\implies \:\:\sf \dfrac{(Cos\: A+ Sin A)\cancel{(CosA - Sin A )} }{ \cancel{(CosA-SinA)}}   \\ \\\\

\qquad:\implies \:\bf{L.H.S}\:\bf (Cos\: A+ Sin A)   \\ \\\\

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Therefore,

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

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

As , We can see that ,

  • \bf{\underline { L.H.S = R.H.S }}\\\\

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

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