हिंदी

Evaluate the Following: `Cosec^-1(Cosec (13pi)/6)` - Mathematics

Advertisements
Advertisements

प्रश्न

Evaluate the following:

`cosec^-1(cosec  (13pi)/6)`

Advertisements

उत्तर

We know that

cosec-1 (cosec θ) = θ,    [-π/2,0) ∪ (0,π/2]

`cosec^-1(cosec  (13pi)/6)=cosec^-1[cosec(2pi+pi/6)]`

`=cosec^-1(cosec  pi/6)`

`=pi/6`

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 4: Inverse Trigonometric Functions - Exercise 4.07 [पृष्ठ ४२]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 12
अध्याय 4 Inverse Trigonometric Functions
Exercise 4.07 | Q 5.5 | पृष्ठ ४२

वीडियो ट्यूटोरियलVIEW ALL [2]

संबंधित प्रश्न

If (tan1x)2 + (cot−1x)2 = 5π2/8, then find x.


Prove that

`tan^(-1) [(sqrt(1+x)-sqrt(1-x))/(sqrt(1+x)+sqrt(1-x))]=pi/4-1/2 cos^(-1)x,-1/sqrt2<=x<=1`


​Find the principal values of the following:

`cos^-1(tan  (3pi)/4)`


`sin^-1(sin  (7pi)/6)`


`sin^-1(sin3)`


`sin^-1(sin12)`


Evaluate the following:

`cos^-1{cos  (5pi)/4}`


Evaluate the following:

`tan^-1(tan  (9pi)/4)`


Evaluate the following:

`tan^-1(tan4)`


Evaluate the following:

`\text(cosec)^-1(\text{cosec}  pi/4)`


Evaluate the following:

`cosec^-1{cosec  (-(9pi)/4)}`


Write the following in the simplest form:

`tan^-1{(sqrt(1+x^2)-1)/x},x !=0`


Evaluate the following:

`sin(cos^-1  5/13)`


Evaluate:

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


If `cot(cos^-1  3/5+sin^-1x)=0`, find the values of x.


`sin^-1x=pi/6+cos^-1x`


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+tan^-1  x/3=pi/4, 0<x<sqrt6`


Solve the following:

`sin^-1x+sin^-1  2x=pi/3`


Evaluate the following:

`tan  1/2(cos^-1  sqrt5/3)`


Evaluate the following:

`sin(2tan^-1  2/3)+cos(tan^-1sqrt3)`


`tan^-1  1/4+tan^-1  2/9=1/2cos^-1  3/2=1/2sin^-1(4/5)`


If `sin^-1  (2a)/(1+a^2)+sin^-1  (2b)/(1+b^2)=2tan^-1x,` Prove that  `x=(a+b)/(1-ab).`


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`


Prove that:

`tan^-1  (2ab)/(a^2-b^2)+tan^-1  (2xy)/(x^2-y^2)=tan^-1  (2alphabeta)/(alpha^2-beta^2),`   where `alpha=ax-by  and  beta=ay+bx.`


If x > 1, then write the value of sin−1 `((2x)/(1+x^2))` in terms of tan−1 x.


Write the value of cos−1 (cos 6).


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


Write the value of  `cot^-1(-x)`  for all `x in R` in terms of `cot^-1(x)`


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

 

If \[\cos\left( \sin^{- 1} \frac{2}{5} + \cos^{- 1} x \right) = 0\], find the value of x.

 

The number of solutions of the equation \[\tan^{- 1} 2x + \tan^{- 1} 3x = \frac{\pi}{4}\] is

 


If \[\cos^{- 1} \frac{x}{2} + \cos^{- 1} \frac{y}{3} = \theta,\]  then 9x2 − 12xy cos θ + 4y2 is equal to


If \[\sin^{- 1} \left( \frac{2a}{1 - a^2} \right) + \cos^{- 1} \left( \frac{1 - a^2}{1 + a^2} \right) = \tan^{- 1} \left( \frac{2x}{1 - x^2} \right),\text{ where }a, x \in \left( 0, 1 \right)\] , then, the value of x is

 


Prove that : \[\cot^{- 1} \frac{\sqrt{1 + \sin x} + \sqrt{1 - \sin x}}{\sqrt{1 + \sin x} - \sqrt{1 - \sin x}} = \frac{x}{2}, 0 < x < \frac{\pi}{2}\] .


The equation sin-1 x – cos-1 x = cos-1 `(sqrt3/2)` has ____________.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×