मराठी

If 3 tan–1x + cot–1x = π, then x equals ______.

Advertisements
Advertisements

प्रश्न

If 3 tan–1x + cot–1x = π, then x equals ______.

पर्याय

  • 0

  • 1

  • – 1

  • `1/2`

MCQ
रिकाम्या जागा भरा
Advertisements

उत्तर

If 3 tan–1x + cot–1x = π, then x equals 1.

Explanation:

Given that 3 tan–1x + cot–1x = π

⇒ 2 tan–1x + tan–1x + cot–1x = π

⇒ `2 tan^-1x + pi/2` = π  ......`[because tan^-1x + cot^-1x = pi/2]`

⇒ `2tan^-1x = pi - pi/2`

⇒ `2tan^-1x = pi/2`

⇒ `2tan^-1x = pi/4`

⇒ `tan^-1x = tan^-1(1)`

⇒ x = 1

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 2: Inverse Trigonometric Functions - Exercise [पृष्ठ ३७]

APPEARS IN

एनसीईआरटी एक्झांप्लर Mathematics Exemplar [English] Class 12
पाठ 2 Inverse Trigonometric Functions
Exercise | Q 22 | पृष्ठ ३७

संबंधित प्रश्‍न

 

Prove that:

`tan^(-1)""1/5+tan^(-1)""1/7+tan^(-1)""1/3+tan^(-1)""1/8=pi/4`

 

If a line makes angles 90°, 60° and θ with x, y and z-axis respectively, where θ is acute, then find θ.


Prove the following: 

3cos1x = cos–1 (4x3 – 3x), `x ∈ [1/2, 1]`


Write the following function in the simplest form:

`tan^(-1) (sqrt((1 - cos x)/(1 + cos x)))`, 0 < x < π


Find the value of the following:

`tan  1/2 [sin^(-1)  (2x)/(1 + x^2) + cos^(-1)  (1 - y^2)/(1 + y^2)], |x| < 1, y > 0 and xy < 1`


Solve for x : `tan^-1 ((2-"x")/(2+"x")) = (1)/(2)tan^-1  ("x")/(2), "x">0.`


Solve: `cot^-1 x - cot^-1 (x + 2) = pi/12, x > 0`


Choose the correct alternative:

If `sin^-1x + sin^-1y = (2pi)/3` ; then `cos^-1x + cos^-1y` is equal to


Choose the correct alternative:

`tan^-1 (1/4) + tan^-1 (2/9)` is equal to


Choose the correct alternative:

`sin^-1 (tan  pi/4) - sin^-1 (sqrt(3/x)) = pi/6`. Then x is a root of the equation


Choose the correct alternative:

sin–1(2 cos2x – 1) + cos1(1 – 2 sin2x) =


Choose the correct alternative:

If `cot^-1(sqrt(sin alpha)) + tan^-1(sqrt(sin alpha))` = u, then cos 2u is equal to


Choose the correct alternative:

If `sin^-1x + cot^-1 (1/2) = pi/2`, then x is equal to


Evaluate tan (tan–1(– 4)).


Prove that cot–17 + cot–18 + cot–118 = cot–13


Evaluate `cos[cos^-1 ((-sqrt(3))/2) + pi/6]`


The value of cot–1(–x) for all x ∈ R in terms of cot–1x is ______.


The minimum value of sinx - cosx is ____________.


The value of `"tan"^ -1 (3/4) + "tan"^-1 (1/7)` is ____________.


If `"tan"^-1 ("cot"  theta) = 2theta, "then"  theta` is equal to ____________.


The domain of the function defind by f(x) `= "sin"^-1 sqrt("x" - 1)` is ____________.


The value of sin (2tan-1 (0.75)) is equal to ____________.


`"cot" ("cosec"^-1  5/3 + "tan"^-1  2/3) =` ____________.


`"tan" (pi/4 + 1/2 "cos"^-1 "x") + "tan" (pi/4 - 1/2 "cos"^-1 "x") =` ____________.


`"cos"^-1 1/2 + 2  "sin"^-1 1/2` is equal to ____________.


`"tan"^-1 (sqrt3)`


Solve for x : `{"x cos" ("cot"^-1 "x") + "sin" ("cot"^-1 "x")}^2` = `51/50


The Government of India is planning to fix a hoarding board at the face of a building on the road of a busy market for awareness on COVID-19 protocol. Ram, Robert and Rahim are the three engineers who are working on this project. “A” is considered to be a person viewing the hoarding board 20 metres away from the building, standing at the edge of a pathway nearby. Ram, Robert and Rahim suggested to the firm to place the hoarding board at three different locations namely C, D and E. “C” is at the height of 10 metres from the ground level. For viewer A, the angle of elevation of “D” is double the angle of elevation of “C” The angle of elevation of “E” is triple the angle of elevation of “C” for the same viewer. Look at the figure given and based on the above information answer the following:

Domain and Range of tan-1 x = ________.


`tan^-1  1/2 + tan^-1  2/11` is equal to


The Simplest form of `cot^-1 (1/sqrt(x^2 - 1))`, |x| > 1 is


Find the value of `tan^-1 [2 cos (2 sin^-1  1/2)] + tan^-1 1`.


If `tan^-1 ((x - 1)/(x + 1)) + tan^-1 ((2x - 1)/(2x + 1)) = tan^-1 (23/36)` = then prove that 24x2 – 23x – 12 = 0


`tan^-1 sqrt3 - cot^-1 (- sqrt3)` is equal to ______.


Solve:

sin–1 (x) + sin–1 (1 – x) = cos–1 x


If \[\tan^{-1}\left(\frac{x}{2}\right)+\tan^{-1}\left(\frac{y}{2}\right)+\tan^{-1}\left(\frac{z}{2}\right)=\frac{\pi}{2}\]  then xy + yz + zx =


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×