English

Solve the Following Equation For X: `Tan^-1(2+X)+Tan^-1(2-x)=Tan^-1 2/3, Where X< -sqrt3 Or, X>Sqrt3` - Mathematics

Advertisements
Advertisements

Question

Solve the following equation for x:

`tan^-1(2+x)+tan^-1(2-x)=tan^-1  2/3, where  x< -sqrt3 or, x>sqrt3`

Advertisements

Solution

We know

`tan^-1x+tan^-1y=tan^-1((x+y)/(1-xy))`

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

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

⇒ `4/(1-4+x^2)=2/3`

⇒ `-6+2x^2=12`

⇒ `2x^2=18`

⇒ `x^2=9`

⇒ `x=+-3`

shaalaa.com
  Is there an error in this question or solution?
Chapter 4: Inverse Trigonometric Functions - Exercise 4.11 [Page 82]

APPEARS IN

RD Sharma Mathematics [English] Class 12
Chapter 4 Inverse Trigonometric Functions
Exercise 4.11 | Q 3.09 | Page 82

RELATED QUESTIONS

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


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


If tan-1x+tan-1y=π/4,xy<1, then write the value of x+y+xy.


`sin^-1(sin  (7pi)/6)`


Evaluate the following:

`cos^-1{cos  (13pi)/6}`


Evaluate the following:

`tan^-1(tan2)`


Evaluate the following:

`sec^-1(sec  (9pi)/5)`


Evaluate the following:

`sec^-1(sec  (13pi)/4)`


Evaluate the following:

`cot^-1(cot  pi/3)`


Evaluate the following:

`cot^-1(cot  (19pi)/6)`


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 the following:

`sin(tan^-1  24/7)`


Prove the following result-

`tan^-1  63/16 = sin^-1  5/13 + cos^-1  3/5`


Evaluate:

`cos{sin^-1(-7/25)}`


Evaluate:

`tan{cos^-1(-7/25)}`


Evaluate:

`sin(tan^-1x+tan^-1  1/x)` for x > 0


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


`(9pi)/8-9/4sin^-1  1/3=9/4sin^-1  (2sqrt2)/3`


`2tan^-1(1/2)+tan^-1(1/7)=tan^-1(31/17)`


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`


What is the value of cos−1 `(cos  (2x)/3)+sin^-1(sin  (2x)/3)?`


Write the range of tan−1 x.


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


Write the value of cos\[\left( \frac{1}{2} \cos^{- 1} \frac{3}{5} \right)\]


Write the value of tan1\[\left\{ \tan\left( \frac{15\pi}{4} \right) \right\}\]


Show that \[\sin^{- 1} (2x\sqrt{1 - x^2}) = 2 \sin^{- 1} x\]


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


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

 

2 tan−1 {cosec (tan−1 x) − tan (cot1 x)} is equal to


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


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


It \[\tan^{- 1} \frac{x + 1}{x - 1} + \tan^{- 1} \frac{x - 1}{x} = \tan^{- 1}\]   (−7), then the value of x is

 


Find : \[\int\frac{2 \cos x}{\left( 1 - \sin x \right) \left( 1 + \sin^2 x \right)}dx\] .


Find the value of x, if tan `[sec^(-1) (1/x) ] = sin ( tan^(-1) 2) , x > 0 `.


Find the real solutions of the equation
`tan^-1 sqrt(x(x + 1)) + sin^-1 sqrt(x^2 + x + 1) = pi/2`


The value of sin `["cos"^-1 (7/25)]` is ____________.


Find the value of `sin^-1(cos((33π)/5))`.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×