Question

If 2sinβ + 3 cos β =2 , prove that ( 3 sin β – 2cos β) = ± 3​

Answer 1

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

Given Trigonometric function is

$\rm :\longmapsto\: 2sin\beta + 3cos\beta = 2$

Now, Consider

$\rm :\longmapsto\: {(2sin\beta + 3cos\beta )}^{2} + {(3sin\beta - 2cos\beta )}^{2}$

We know,

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

and

$\boxed{ \tt{ \: {(x - y)}^{2} = {x}^{2} + {y}^{2} - 2xy \: }}$

So, using these Identities, we get

$\rm \:  =  4 {sin}^{2}\beta + {9cos}^{2}\beta + 12sin\beta cos\beta + {9sin}^{2}\beta + {4cos}^{2}\beta - 12sin\beta cos\beta$

$\rm \:  =  \: {13sin}^{2}\beta + {13cos}^{2}\beta$

$\rm \:  =  \:13( {sin}^{2}\beta + {cos}^{2}\beta )$

$\rm \:  =  \:13 \times 1$

$\rm \:  =  \:13$

Thus,

$\rm :\longmapsto\: {(2sin\beta + 3cos\beta )}^{2} + {(3sin\beta - 2cos\beta )}^{2} = 13$

$\rm :\longmapsto\: {(2)}^{2} + {(3sin\beta - 2cos\beta )}^{2} = 13$

$\rm :\longmapsto\: 4 + {(3sin\beta - 2cos\beta )}^{2} = 13$

$\rm :\longmapsto\: {(3sin\beta - 2cos\beta )}^{2} = 13 - 4$

$\rm :\longmapsto\: {(3sin\beta - 2cos\beta )}^{2} = 9$

$\bf\implies \:3sin\beta - 2cos\beta = \: \pm \: 3$

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

Answer 2

Answer:

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

Given Trigonometric function is

$\rm :\longmapsto\: 2sin\beta + 3cos\beta = 2$

Now, Consider

$\rm :\longmapsto\: {(2sin\beta + 3cos\beta )}^{2} + {(3sin\beta - 2cos\beta )}^{2}$

We know,

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

and

$\boxed{ \tt{ \: {(x - y)}^{2} = {x}^{2} + {y}^{2} - 2xy \: }}$

So, using these Identities, we get

$\rm \:  =  4 {sin}^{2}\beta + {9cos}^{2}\beta + 12sin\beta cos\beta + {9sin}^{2}\beta + {4cos}^{2}\beta - 12sin\beta cos\beta$

$\rm \:  =  \: {13sin}^{2}\beta + {13cos}^{2}\beta$

$\rm \:  =  \:13( {sin}^{2}\beta + {cos}^{2}\beta )$

$\rm \:  =  \:13 \times 1$

$\rm \:  =  \:13$

Thus,

$\rm :\longmapsto\: {(2sin\beta + 3cos\beta )}^{2} + {(3sin\beta - 2cos\beta )}^{2} = 13$

$\rm :\longmapsto\: {(2)}^{2} + {(3sin\beta - 2cos\beta )}^{2} = 13$

$\rm :\longmapsto\: 4 + {(3sin\beta - 2cos\beta )}^{2} = 13$

$\rm :\longmapsto\: {(3sin\beta - 2cos\beta )}^{2} = 13 - 4$

$\rm :\longmapsto\: {(3sin\beta - 2cos\beta )}^{2} = 9$

$\bf\implies \:3sin\beta - 2cos\beta = \: \pm \: 3$

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