Question

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

Answer 1

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