Question

If 1+ sin² α = 3 sin α cos α, then values of cot α are a) -1, 1 b) 0,1 c) 1, 2 d) -1,-1​

Answer 1

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

Given Trigonometric equation is

$\rm :\longmapsto\:1 + {sin}^{2} \alpha = 3sin \alpha cos \alpha$

Divide both sides by sin² α, we get

$\rm :\longmapsto\:\dfrac{1 + {sin}^{2} \alpha }{ {sin}^{2} \alpha } = \dfrac{3sin \alpha cos \alpha }{ {sin}^{2} \alpha }$

$\rm :\longmapsto\:\dfrac{1}{ {sin}^{2} \alpha } + 1 = \dfrac{3cos \alpha }{sin \alpha }$

$\rm :\longmapsto\: {cosec}^{2} \alpha + 1 = 3cot \alpha$

We know,

$\purple{\rm :\longmapsto\:\boxed{\tt{ {cosec}^{2}x - {cot}^{2}x = 1 \: }}}$

So, using this, we get

$\rm :\longmapsto\: {cot}^{2} \alpha + 1 + 1 = 3cot \alpha$

$\rm :\longmapsto\: {cot}^{2} \alpha + 2 = 3cot \alpha$

$\rm :\longmapsto\: {cot}^{2} \alpha - 3cot \alpha + 2 = 0$

$\rm :\longmapsto\: {cot}^{2} \alpha - 2cot \alpha - cot \alpha + 2 = 0$

$\rm :\longmapsto\:cot \alpha (cot \alpha - 2) - 1(cot \alpha - 2) = 0$

$\rm :\longmapsto\: (cot \alpha - 2)(cot \alpha - 1) = 0$

$\rm\implies \:cot \alpha = 2 \: \: or \: \: cot \alpha = 1$

• Hence, option (c) is correct

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

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:

$\large\underline{\bf{Question-}}$

If 1+ sin² α = 3 sin α cos α, then values of cot α are

a) -1, 1

b) 0,1

c) 1, 2

d) -1,-1

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

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

Given Trigonometric equation is

$\rm :\longmapsto\:1 + {sin}^{2} \alpha = 3sin \alpha cos \alpha$

Divide both sides by sin² α, we get

$\rm :\longmapsto\:\dfrac{1 + {sin}^{2} \alpha }{ {sin}^{2} \alpha } = \dfrac{3sin \alpha cos \alpha }{ {sin}^{2} \alpha }$

$\rm :\longmapsto\:\dfrac{1}{ {sin}^{2} \alpha } + 1 = \dfrac{3cos \alpha }{sin \alpha }$

$\rm :\longmapsto\: {cosec}^{2} \alpha + 1 = 3cot \alpha$

We know,

$\purple{\rm :\longmapsto\:\boxed{\tt{ {cosec}^{2}x - {cot}^{2}x = 1 \: }}}$

So, using this, we get

$\rm :\longmapsto\: {cot}^{2} \alpha + 1 + 1 = 3cot \alpha$

$\rm :\longmapsto\: {cot}^{2} \alpha + 2 = 3cot \alpha$

$\rm :\longmapsto\: {cot}^{2} \alpha - 3cot \alpha + 2 = 0$

$\rm :\longmapsto\: {cot}^{2} \alpha - 2cot \alpha - cot \alpha + 2 = 0$

$\rm :\longmapsto\:cot \alpha (cot \alpha - 2) - 1(cot \alpha - 2) = 0$

$\rm :\longmapsto\: (cot \alpha - 2)(cot \alpha - 1) = 0$

$\rm\implies \:cot \alpha = 2 \: \: or \: \: cot \alpha = 1$

Hence, option (c) is correct

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬