Question

Da 4) 1;) It 3 sinA + 5 cos A =5, then show that S sinA - 3 cos A=+-3​

Answer 1

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

Given that,

$\rm :\longmapsto\:3sinA + 5cosA = 5$

On squaring both sides, we get

$\rm :\longmapsto\:(3sinA + 5cosA) {}^{2} = 25$

We know,

$\red{\boxed{ \rm{ \: {(x + y)}^{2} = {x}^{2} + {y}^{2} + 2xy}}}$

So, using this identity, we get

$\rm :\longmapsto\: {(3sinA)}^{2} + {(5cosA)}^{2} + 2(3sinA)(5cosA) = 25$

$\rm :\longmapsto\: 9{sin}^{2}A + 25{cos}^{2}A + 30sinAcosA = 25$

We know,

$\red{\boxed{ \rm{ \: {sin}^{2}x + {cos}^{2}x = 1}}}$

So, using this, we get

$\rm :\longmapsto\: 9(1 - {cos}^{2}A) + 25(1 - {sin}^{2}A) + 30sinAcosA = 25$

$\rm :\longmapsto\: 9 - 9{cos}^{2}A + 25 - 2{sin}^{2}A + 30sinAcosA = 25$

$\rm :\longmapsto\: - 9{cos}^{2}A - 25{sin}^{2}A + 30sinAcosA = - 9$

$\rm :\longmapsto\: 9{cos}^{2}A + 25{sin}^{2}A - 30sinAcosA = 9$

$\rm :\longmapsto\: {(3cosA)}^{2}+ {(5sinA)}^{2} - 30sinAcosA = 9$

$\rm :\longmapsto\: {(3cosA)}^{2}+ {(5sinA)}^{2} - 2(5sinA)(3cosA) = 9$

$\rm :\longmapsto\:(5sinA - 3cosA)^{2} = 9$

$\red{\bf\implies \: \red{\boxed{ \rm{ \: 5sinA - 3cosA = \: \pm \: 3 \: \: }}}}$

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

Answer 2

Answer:

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

Given that,

$\rm :\longmapsto\:3sinA + 5cosA = 5$

On squaring both sides, we get

$\rm :\longmapsto\:(3sinA + 5cosA) {}^{2} = 25$

We know,

$\red{\boxed{ \rm{ \: {(x + y)}^{2} = {x}^{2} + {y}^{2} + 2xy}}}$

So, using this identity, we get

$\rm :\longmapsto\: {(3sinA)}^{2} + {(5cosA)}^{2} + 2(3sinA)(5cosA) = 25$

$\rm :\longmapsto\: 9{sin}^{2}A + 25{cos}^{2}A + 30sinAcosA = 25$

We know,

$\red{\boxed{ \rm{ \: {sin}^{2}x + {cos}^{2}x = 1}}}$

So, using this, we get

$\rm :\longmapsto\: 9(1 - {cos}^{2}A) + 25(1 - {sin}^{2}A) + 30sinAcosA = 25$

$\rm :\longmapsto\: 9 - 9{cos}^{2}A + 25 - 2{sin}^{2}A + 30sinAcosA = 25$

$\rm :\longmapsto\: - 9{cos}^{2}A - 25{sin}^{2}A + 30sinAcosA = - 9$

$\rm :\longmapsto\: 9{cos}^{2}A + 25{sin}^{2}A - 30sinAcosA = 9$

$\rm :\longmapsto\: {(3cosA)}^{2}+ {(5sinA)}^{2} - 30sinAcosA = 9$

$\rm :\longmapsto\: {(3cosA)}^{2}+ {(5sinA)}^{2} - 2(5sinA)(3cosA) = 9$

$\rm :\longmapsto\:(5sinA - 3cosA)^{2} = 9$

$\red{\bf\implies \: \red{\boxed{ \rm{ \: 5sinA - 3cosA = \: \pm \: 3 \: \: }}}}$

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