Question

If $1 + \sin {}^2 \beta = 3 \sin \beta \cos$, then the values of $\cot \beta$ can be:-
(1) 1, - 1
(2) 0, 1
(3) 1, 2
(4) - 1, - 1 ​

Answer 1

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

Given that,

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

can be rewritten as

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

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

We know,

$\boxed{ \tt{ \: \frac{1}{sinx} = cosecx \: }} \\ \\ and \\ \\ \boxed{ \tt{ \: \frac{cosx}{sinx} = cotx \: }}$

So, using these, we get

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

can be rewritten as

$\rm :\longmapsto\: ({cot}^{2}\beta + 1) + 1 = 3cot\beta$

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

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

Now, its a quadratic, so using Splitting of middle terms, we have

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

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

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

$\rm \implies\:\boxed{ \tt{ \: cot\beta = 1 \: \: or \: \: cot\beta = 2 \: }}$

So, Option (3) is correct.

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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:

option [3] 1,2 is the answer