cosecθ⋅cotθ=2√3
cscθ cotθ = 2√3
checking the easy stuff first, you know that
csc(π/6) = 2 and cot(π/6) = √3
so, clearly θ = π/6 is one solution
θ = -π/6 also works, since there both csc and cot are negative
Or, if you want to go the algebraic route,
cosθ/sin^2θ = 2√3
cosθ = 2√3 (1-cos^2θ)
2√3 cos^2θ + cosθ - 2√3 = 0
cosθ = (-1±√(1+48))/(4√3) = (-1±7)/(4√3)
cosθ = √3/2 or -2/√3
θ = π/6
I suppose we are solving for θ
cosecθ⋅cotθ=2√3
(1/sinθ)(cos/sin) = 2√3
cosθ/sin^2θ = 2√3
2√3 sin^2 θ = cosθ
2√3(1-cos^2 θ) = cosθ
2√3cos^2 θ + cosθ - 2√3 = 0
cosθ = (-1 ± √(1 - 4(2√3)(2√3))/(4√3)
= (-1 ± √49)/(4√3)
= 3/(2√3) or - 2/√3
if cosθ = 3/(2√3), θ = 30° or 330°
if cosθ = -2/√3 <----- not possible since -1 < cosθ < 1
θ = 30° or 330° or θ = π/6 or 11π/6
csc θ ⋅ cot θ = 2√3
Subtract 2√3 to both sides
csc θ ⋅ cot θ - 2√3 = 0
1 / sin θ ⋅ cos θ / sin θ - 2√3 = 0
cos θ / sin² θ - 2√3 = 0
cos θ / sin² θ - 2√3 ⋅ sin² θ / sin² θ = 0
( cos θ - 2√3 ⋅ sin² θ ) / sin² θ = 0
Fraction = 0 when numerator = 0
So you must solve:
cos θ - 2√3 ⋅ sin² θ = 0
Use identity:
sin² θ = 1 - cos² θ
cos θ - 2√3 ⋅ ( 1 - cos² θ ) = 0
cos θ - 2√3 + 2√3 ⋅ cos² θ = 0
2√3 ⋅ cos² θ + cos θ - 2√3 + = 0
Use substitution:
cos θ = x
2√3 ⋅ x² + x - 2√3 = 0
Use quadratic formula:
x1/2= [ - b ± √ ( b² - 4 ⋅ a ⋅ c ) ] / 2 ⋅ a
In this case:
a = 2√3 , b = 1 , c = - 2√3
x1/2= [ - 1 ± √ 1² - 4 ⋅ 2 √3 ⋅ ( - 2√3 ) ] / 2 ⋅ 2√3 =
[ - 1 ± √ ( 1 + 8 √3 ⋅ 2√3 ) ] / 4√3 =
[ - 1 ± √ ( 1 + 16 √3 ⋅√3 ) ] / 4√3 =
[ - 1 ± √ ( 1 + 16 ⋅ 3 ) ] / 4√3 =
[ - 1 ± √ ( 1 + 48 ) ] / 4√3 =
( - 1 ± √ 49 ) / 4√3 =
( - 1 ± 7 ) / 4√3
The solutions are:
x = ( - 1 - 7 ) / 4√3 = - 8 / 4√3 = - 4 ⋅ 2 / 4√3 = - 2 / √3
and
x = ( - 1 + 7 ) / 4√3 = 6 / 4√3 = 2 ⋅ 3 / 2 ⋅ 2 ⋅ √3 = 3 / 2 √3 = √3 ⋅ √3 / 2 √3 = √3 / 2
Substitute back :
cos θ = x
The solutions are:
cos θ = - 2 √3 / 3 and cos θ = √3 / 2
Solution cos θ = - 2 √3 / 3 = -1.1547005 you must discard because cosine can be - 1 ÷ 1.
So solution is:
cos θ = √3 / 2
In interval 0 to 2 π for angles θ = π / 6 and θ = 11 π / 6 , cos θ = √3 / 2
General solutions for cos θ = √3 / 2 :
θ = π / 6 ± 2 π n
and
θ = 11 π / 6 ± 2 π n
Because period of cos θ = 2 π
n = 1 , 2 , 3 ....
To find the value of θ that satisfies the equation cosecθ⋅cotθ=2√3, we need to understand the definitions of cosecant (cosec) and cotangent (cot) and their relationship with sine (sin) and cosine (cos).
The cosecant of an angle θ, denoted as cosecθ, is the reciprocal of the sine of θ. In formula form, cosecθ = 1/sinθ.
The cotangent of an angle θ, denoted as cotθ, is the reciprocal of the tangent of θ. In formula form, cotθ = 1/tanθ.
Using these definitions, we can rewrite the given equation as:
(1/sinθ) * (1/tanθ) = 2√3
To simplify the equation, we need to express cotθ in terms of sinθ. Since cotθ = 1/tanθ, and tanθ = sinθ/cosθ, we can substitute this value in:
(1/sinθ) * (1 / (sinθ / cosθ)) = 2√3
Simplifying further, we get:
(1/sinθ) * (cosθ / sinθ) = 2√3
When we multiply these terms, the sinθ in the numerator and denominator cancel out, leaving:
cosθ = 2√3
So, the equation becomes:
cosθ = 2√3
To find the value of θ, we need to take the inverse cosine (cos⁻¹) of both sides of the equation:
θ = cos⁻¹(2√3)
Using a calculator, we can determine the value of θ to be approximately 30 degrees or π/6 radians.