English

Evaluate the Following: `Cot^-1(Cot (19pi)/6)` - Mathematics

Advertisements
Advertisements

Question

Evaluate the following:

`cot^-1(cot  (19pi)/6)`

Advertisements

Solution

We know that

cot-1 (cot θ) = θ,   (0, π)

We have

`cot^-1(cot  (19pi)/6)=cot^-1[cot(pi+pi/6)]`

`=cot^-1(cot  pi/6)`

`=pi/6`

shaalaa.com
  Is there an error in this question or solution?
Chapter 4: Inverse Trigonometric Functions - Exercise 4.07 [Page 43]

APPEARS IN

RD Sharma Mathematics [English] Class 12
Chapter 4 Inverse Trigonometric Functions
Exercise 4.07 | Q 6.4 | Page 43

RELATED QUESTIONS

If sin [cot−1 (x+1)] = cos(tan1x), then find x.


`sin^-1(sin4)`


Evaluate the following:

`cos^-1(cos5)`


Evaluate the following:

`tan^-1(tan4)`


Evaluate the following:

`sec^-1(sec  (7pi)/3)`


Evaluate the following:

`cosec^-1(cosec  (3pi)/4)`


Evaluate the following:

`cosec^-1(cosec  (13pi)/6)`


Evaluate the following:

`cosec^-1{cosec  (-(9pi)/4)}`


Write the following in the simplest form:

`tan^-1{x+sqrt(1+x^2)},x in R `


Write the following in the simplest form:

`tan^-1{sqrt(1+x^2)-x},x in R`


Evaluate:

`cosec{cot^-1(-12/5)}`


If `cos^-1x + cos^-1y =pi/4,`  find the value of `sin^-1x+sin^-1y`


If `sin^-1x+sin^-1y=pi/3`  and  `cos^-1x-cos^-1y=pi/6`,  find the values of x and y.


Solve the following equation for x:

`tan^-1((1-x)/(1+x))-1/2 tan^-1x` = 0, where x > 0


Solve the following equation for x:

 cot−1x − cot−1(x + 2) =`pi/12`, > 0


Solve the following equation for x:

`tan^-1  (x-2)/(x-1)+tan^-1  (x+2)/(x+1)=pi/4`


Evaluate: `cos(sin^-1  3/5+sin^-1  5/13)`


Solve the following:

`cos^-1x+sin^-1  x/2=π/6`


`2sin^-1  3/5-tan^-1  17/31=pi/4`


Solve the following equation for x:

`tan^-1  1/4+2tan^-1  1/5+tan^-1  1/6+tan^-1  1/x=pi/4`


Solve the following equation for x:

`2tan^-1(sinx)=tan^-1(2sinx),x!=pi/2`


Solve the following equation for x:

`cos^-1((x^2-1)/(x^2+1))+1/2tan^-1((2x)/(1-x^2))=(2x)/3`


Solve the following equation for x:

`tan^-1((x-2)/(x-1))+tan^-1((x+2)/(x+1))=pi/4`


If `sin^-1x+sin^-1y+sin^-1z=(3pi)/2,`  then write the value of x + y + z.


Write the value of sin (cot−1 x).


Write the value of sin1 (sin 1550°).


If \[\tan^{- 1} (\sqrt{3}) + \cot^{- 1} x = \frac{\pi}{2},\] find x.


If x < 0, y < 0 such that xy = 1, then write the value of tan1 x + tan−1 y.


Write the value of \[\tan^{- 1} \left\{ 2\sin\left( 2 \cos^{- 1} \frac{\sqrt{3}}{2} \right) \right\}\]


Wnte the value of the expression \[\tan\left( \frac{\sin^{- 1} x + \cos^{- 1} x}{2} \right), \text { when } x = \frac{\sqrt{3}}{2}\]


If \[\cos\left( \tan^{- 1} x + \cot^{- 1} \sqrt{3} \right) = 0\] , find the value of x.

 

Find the value of \[2 \sec^{- 1} 2 + \sin^{- 1} \left( \frac{1}{2} \right)\]


The positive integral solution of the equation
\[\tan^{- 1} x + \cos^{- 1} \frac{y}{\sqrt{1 + y^2}} = \sin^{- 1} \frac{3}{\sqrt{10}}\text{ is }\]


If sin−1 − cos−1 x = `pi/6` , then x = 


If 4 cos−1 x + sin−1 x = π, then the value of x is

 


Prove that : \[\tan^{- 1} \left( \frac{\sqrt{1 + x^2} + \sqrt{1 - x^2}}{\sqrt{1 + x^2} - \sqrt{1 - x^2}} \right) = \frac{\pi}{4} + \frac{1}{2} \cos^{- 1} x^2 ;  1 < x < 1\].


If \[\tan^{- 1} \left( \frac{1}{1 + 1 . 2} \right) + \tan^{- 1} \left( \frac{1}{1 + 2 . 3} \right) + . . . + \tan^{- 1} \left( \frac{1}{1 + n . \left( n + 1 \right)} \right) = \tan^{- 1} \theta\] , then find the value of θ.


Solve for x : {xcos(cot-1 x) + sin(cot-1 x)}= `51/50`


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×