मराठी

Solve the Following Equation For X: Cot−1x − Cot−1(X + 2) =`Pi/12`, X > 0 - Mathematics

Advertisements
Advertisements

प्रश्न

Solve the following equation for x:

 cot−1x − cot−1(x + 2) =`pi/12`, > 0

Advertisements

उत्तर

⇒ `cot^-1(x)-cot^-1(x+2)=pi/12`

⇒ `tan^-1(1/x)+cot^-1(1/(x+2))=pi/12`     `[because cot^-1x=tan^-1 1/x]`

⇒ `tan^-1((1/x-1/(x+2))/(1+1/(x(x+2))))=pi/12`

⇒ `tan^-1((2/(x(x+2)))/((x^2+2x+1)/(x(x+2))))=pi/12`

⇒ `tan^-1(2/(x^2+2x+1))=pi/12`

⇒ `(2/(x^2+2x+1))=tan  pi/12`

⇒ `(2/(x^2+2x+1))=tan(pi/3-pi/4)`

⇒ `(2/(x^2+2x+1))=(tan  pi/3-tan  pi/4)/(1+tan  pi/3xxtan  pi/4`

⇒ `(2/(x^2+2x+1))=(sqrt3-1)/(sqrt3+1)`

⇒ `(2/(x^2+2x+1))=(sqrt3-1)/(sqrt3+1)xx(sqrt3+1)/(sqrt3+1)`

⇒ `(2/(x^2+2x+1))=2/(sqrt3+1)^2`

⇒ `1/(x+1)^2=1/(sqrt3+1)^2`

⇒ `x+1=sqrt3+1`

⇒ `x=sqrt3`

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 4: Inverse Trigonometric Functions - Exercise 4.11 [पृष्ठ ८२]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 12
पाठ 4 Inverse Trigonometric Functions
Exercise 4.11 | Q 3.05 | पृष्ठ ८२

व्हिडिओ ट्यूटोरियलVIEW ALL [2]

संबंधित प्रश्‍न

Write the value of `tan(2tan^(-1)(1/5))`


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`


`sin^-1(sin2)`


Evaluate the following:

`tan^-1(tan  (9pi)/4)`


Evaluate the following:

`tan^-1(tan12)`


Evaluate the following:

`sec^-1(sec  pi/3)`


Evaluate the following:

`sec^-1(sec  (25pi)/6)`


Write the following in the simplest form:

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


Prove the following result

`tan(cos^-1  4/5+tan^-1  2/3)=17/6`


Evaluate:

`cot{sec^-1(-13/5)}`


Evaluate:

`cos(tan^-1  3/4)`


Evaluate: `sin{cos^-1(-3/5)+cot^-1(-5/12)}`


`4sin^-1x=pi-cos^-1x`


Solve the following equation for x:

`tan^-1  2x+tan^-1  3x = npi+(3pi)/4`


Solve the following equation for x:

`tan^-1  x/2+tan^-1  x/3=pi/4, 0<x<sqrt6`


Solve the following equation for x:

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


Evaluate the following:

`tan  1/2(cos^-1  sqrt5/3)`


Solve the following equation for x:

`3sin^-1  (2x)/(1+x^2)-4cos^-1  (1-x^2)/(1+x^2)+2tan^-1  (2x)/(1-x^2)=pi/3`


Write the value of sin (cot−1 x).


Write the range of tan−1 x.


Write the value of sin1 (sin 1550°).


Write the value of cos−1 \[\left( \tan\frac{3\pi}{4} \right)\]


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


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


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


\[\text{ If } u = \cot^{- 1} \sqrt{\tan \theta} - \tan^{- 1} \sqrt{\tan \theta}\text{ then }, \tan\left( \frac{\pi}{4} - \frac{u}{2} \right) =\]


\[\text{ If }\cos^{- 1} \frac{x}{3} + \cos^{- 1} \frac{y}{2} = \frac{\theta}{2}, \text{ then }4 x^2 - 12xy \cos\frac{\theta}{2} + 9 y^2 =\]


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


If tan−1 3 + tan−1 x = tan−1 8, then x =


If θ = sin−1 {sin (−600°)}, then one of the possible values of θ is

 


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

 


If 2 tan−1 (cos θ) = tan−1 (2 cosec θ), (θ ≠ 0), then find the value of θ.


Find the domain of `sec^(-1) x-tan^(-1)x`


Find the simplified form of `cos^-1 (3/5 cosx + 4/5 sin x)`, where x ∈ `[(-3pi)/4, pi/4]`


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


The value of tan `("cos"^-1  4/5 + "tan"^-1  2/3) =`


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


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×