Math, asked by bhanusonker21, 2 months ago

If sinA=(sinx+siny)/(1+sinxsiny) then proof that cosB=(cosxcosy)/(1+sinxsiny).

Answers

Answered by mathdude500
5

Correct Statement is

 \sf \: If  \: sinA \:  =  \: \dfrac{sinx + siny}{1 + sinxsiny}  \: then \: prove \: cosA = \dfrac{cosxcosy}{1 + sinxsiny}

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

We know that

\rm :\longmapsto\: {sin}^{2} A +  {cos}^{2} A = 1

\rm :\longmapsto\: {cos}^{2} A = 1 -  {sin}^{2} A

↝ On substituting the value of sinA, we get

 \sf \:  \:  \:  \:  =  \:  \: 1 -  {\bigg(\: \dfrac{sinx + siny}{1 + sinxsiny} \bigg) }^{2}

 \sf \:  \:  \:  \:  =  \:  \: 1 - \dfrac{ {sin}^{2}x +  {sin}^{2}y + 2sinxsiny}{1 +  {sin}^{2}x {sin}^{2}y + 2sinxsiny}

 \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \boxed{ \blue{ \bf \because \:  {(x + y)}^{2} =  {x}^{2}   +  {y}^{2}  + 2xy}}

 \sf \:  \:  \:  \:  =  \:  \: \dfrac{ 1 +  {sin}^{2}x {sin}^{2}y + 2sinxsiny - {sin}^{2}x  -   {sin}^{2}y - 2sinxsiny}{1 +  {sin}^{2}x {sin}^{2}y + 2sinxsiny}

 \sf \:  \:  \:  \:  =  \:  \: \dfrac{ 1 + {sin}^{2}x{sin}^{2}y -  {sin}^{2}x -  {sin}^{2}y}{1 +  {sin}^{2}x {sin}^{2}y + 2sinxsiny}

 \sf \:  \:  \:  \:  =  \:  \: \dfrac{(1  -  {sin}^{2}x)  + ({sin}^{2}y{sin}^{2}x - {sin}^{2}y)}{1 +  {sin}^{2}x {sin}^{2}y + 2sinxsiny}

 \sf \:  \:  \:  \:  =  \:  \: \dfrac{ {cos}^{2}x -  {sin}^{2}y(1 -  {sin}^{2}x)}{1 +  {sin}^{2}x {sin}^{2}y + 2sinxsiny}

 \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \boxed{ \bf \because \: 1 -  {sin}^{2}A =  {cos}^{2}A}

 \sf \:  \:  \:  \:  =  \:  \: \dfrac{ {cos}^{2}x -  {sin}^{2}y {cos}^{2}x}{1 +  {sin}^{2}x {sin}^{2}y + 2sinxsiny}

 \sf \:  \:  \:  \:  =  \:  \: \dfrac{ {cos}^{2}x(1 -  {sin}^{2}y)}{1 +  {sin}^{2}x {sin}^{2}y + 2sinxsiny}

 \sf \:  \:  \:  \:  =  \:  \: \dfrac{ {cos}^{2}x {cos}^{2}y}{1 +  {sin}^{2}x {sin}^{2}y + 2sinxsiny}

 \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \boxed{ \bf \because \: 1 -  {sin}^{2}A =  {cos}^{2}A}

 \sf \:  \:  \:  \:  =  \:  \:  {\bigg(\dfrac{cosxcosy}{1 + sinxsiny}  \bigg) }^{2}

\rm :\implies\: \sf \:  \:  \: {cos}^{2}A   \:  =  \:  \:  {\bigg(\dfrac{cosxcosy}{1 + sinxsiny}  \bigg) }^{2}

\rm :\implies\:cosA =  \sf \:  \:  \:   \:  \:  {\bigg(\dfrac{cosxcosy}{1 + sinxsiny}  \bigg) }

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

Additional Information:-

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

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