हिंदी

Find the general solutions of the following equation: tanθ=-3

Advertisements
Advertisements

प्रश्न

Find the general solutions of the following equation:

`tan theta = - sqrt3`

योग
Advertisements

उत्तर

The general solution of tan θ = tan α is

θ = nπ + α, n ∈ Z.

Now, tan θ = - `sqrt3`

∴ tan θ = - `tan  pi/3    ....[because "tan" pi/3 = sqrt3]`

∴ tan θ = tan`(pi - pi/3)   ...[because tan (pi - theta) = - tan theta]`

∴ tan θ = tan`(2pi)/3`

∴ the required general solution is

∴ θ = `"n"pi + (2pi)/3,` n ∈ Z

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 3: Trigonometric Functions - Miscellaneous exercise 3 [पृष्ठ १०९]

APPEARS IN

बालभारती Mathematics and Statistics 1 (Arts and Science) [English] Standard 12 Maharashtra State Board
अध्याय 3 Trigonometric Functions
Miscellaneous exercise 3 | Q 4.1 | पृष्ठ १०९

संबंधित प्रश्न

Find the principal solution of the following equation:

sin θ = `-1/2`


Find the principal solution of the following equation: 

tan θ = – 1


Find the principal solution of the following equation:

`sqrt(3)` cosecθ + 2 = 0 


Find the general solution of the following equation:

sinθ = `1/2`.


Find the general solution of the following equation:

sec θ = `sqrt(2)`.


Find the general solution of the following equation:

cosec θ = - √2.


Find the general solution of the following equation:

sin 2θ = `1/2`


Find the general solution of the following equation:

4 cos2 θ  = 3


Find the general solution of the following equation:

4sin2θ = 1.


Find the general solution of the following equation:

cos 4θ = cos 2θ


In ΔABC, prove that `sin(("B" − "C")/2) = (("b" − "c")/"a")cos  "A"/(2)`.


With the usual notations prove that `2{asin^2  "C"/(2) + "c"sin^2  "A"/(2)}` = a – b + c.


Select the correct option from the given alternatives:

In Δ ABC if ∠A = 45°, ∠B = 30°, then the ratio of its sides are


If `"sin"^-1 4/5 + "cos"^-1 12/13 = "sin"^-1 alpha`, then α = ______.


Select the correct option from the given alternatives:

If tan-1(2x) + tan-1(3x) = `pi/4`, then x = _____


`cos[tan^-1  1/3 + tan^-1  1/2]` = ______


Find the principal solutions of the following equation:

cot θ = 0


Find the general solutions of the following equation:

sin θ - cos θ = 1


If sin-1(1 - x) - 2 sin-1x =  `pi/2`, then find the value of x.


Show that `cot^-1  1/3 - tan^-1  1/3 = cot^-1  3/4`.


Prove the following:

`cos^-1 "x" = pi + tan^-1 (sqrt(1 - "x"^2)/"x")`, if x < 0


If | x | < 1, then prove that

`2 tan^-1 "x" = tan^-1 ("2x"/(1 - "x"^2)) = sin^-1 ("2x"/(1 + "x"^2)) = cos^-1 ((1 - "x"^2)/(1 + "x"^2))`


If x, y, z are positive, then prove that

`tan^-1 (("x - y")/(1 + "xy")) + tan^-1 (("y - z")/(1 + "yz")) + tan^-1 (("z - x")/(1 + "zx")) = 0`


The principal solutions of `sqrt(3)` sec x − 2 = 0 are ______


Find the principal solutions of sin x = `-1/2`


The value of tan 57°- tan 12°- tan 57° tan 12° is ______.


If f(x) = sin-1`(sqrt((1 - x)/2))`, then f'(x) = ?


The principal solutions of cot x = `sqrt3` are ______.


If 2 cos2 θ + 3 cos θ = 2, then permissible value of cos θ is ________.


The value of `tan^-1  1/3 + tan^-1  1/5 + tan^-1  1/7 + tan^-1  1/8` is ______.


If 4 sin-1x + 6 cos-1 x = 3π then x = ______.


The value of θ in (π, 2π) satisfying the equation sin2θ - cos2θ = 1 is ______ 


The general solution of 4sin2 x = 1 is ______.


The general solution of cosec x = `-sqrt2` is ______ 


Find the principal solutions of cot θ = 0


General solution of the equation sin 2x – sin 4x + sin 6x = 0 is ______.


Which of the following equation has no solution?


Principal solutions at the equation sin 2x + cos 2x = 0, where π < x < 2 π are ______.


If a = sin θ + cos θ, b = sin3 θ + cos3 θ, then ______.


The general solution of sin x – 3 sin 2x + sin 3x = cos x – 3 cos 2x + cos 3x is ______.


The general solution of cot 4x = –1 is ______.


If `2sin^-1  3/7` = cos–1β, then find the value of β.


If `sin^-1  4/5 + cos^-1  12/13` = sin–1α, then find the value of α.


If tan θ + sec θ = `sqrt(3)`, find the general value of θ.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×