मराठी

Solve the Following Equation For X: `Tan^-1((2x)/(1-x^2))+Cot^-1((1-x^2)/(2x))=(2pi)/3,X>0` - Mathematics

Advertisements
Advertisements

प्रश्न

Solve the following equation for x:

`tan^-1((2x)/(1-x^2))+cot^-1((1-x^2)/(2x))=(2pi)/3,x>0`

उत्तर

We know

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

`thereforetan^-1((2x)/(1-x^2))+cot^-1((1-x^2)/(2x))=(2pi)/3`

`=>tan^-1((2x)/(1-x^2))+tan^-1((2x)/(1-x^2))=(2pi)/3`    `[becausecot^1x=tan^-1  1/x]`

`=>tan^-1((2x)/(1-x^2))=pi/3`

`=>2tan^-1x=pi/3`     `[because2tan^-1xtan^-1((2x)/(1-x^2))]`

`=>tan^-1x=pi/6`

`=>x=tan  pi/6`

`=>x=1/sqrt3`

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 4: Inverse Trigonometric Functions - Exercise 4.14 [पृष्ठ ११६]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 12
पाठ 4 Inverse Trigonometric Functions
Exercise 4.14 | Q 8.3 | पृष्ठ ११६

व्हिडिओ ट्यूटोरियलVIEW ALL [2]

संबंधित प्रश्‍न

Solve the equation for x:sin1x+sin1(1x)=cos1x


 

Prove that :

`2 tan^-1 (sqrt((a-b)/(a+b))tan(x/2))=cos^-1 ((a cos x+b)/(a+b cosx))`

 

 

If `tan^(-1)((x-2)/(x-4)) +tan^(-1)((x+2)/(x+4))=pi/4` ,find the value of x

 

`sin^-1(sin  (7pi)/6)`


`sin^-1(sin  (13pi)/7)`


`sin^-1(sin3)`


Evaluate the following:

`cos^-1(cos12)`


Evaluate the following:

`tan^-1(tan  (7pi)/6)`


Evaluate the following:

`tan^-1(tan2)`


Evaluate the following:

`sec^-1(sec  (7pi)/3)`


Evaluate the following:

`sec^-1{sec  (-(7pi)/3)}`


Evaluate the following:

`cosec^-1(cosec  (11pi)/6)`


Evaluate the following:

`cot^-1(cot  (4pi)/3)`


Evaluate the following:

`cot^-1{cot  ((21pi)/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)-1)/x},x !=0`


Evaluate the following:

`cosec(cos^-1  3/5)`


Evaluate the following:

`cos(tan^-1  24/7)`


Prove the following result-

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


Evaluate:

`cot(tan^-1a+cot^-1a)`


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


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


Solve the following equation for x:

tan−1(x + 2) + tan−1(x − 2) = tan−1 `(8/79)`, x > 0


Solve the following equation for x:

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


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


Solve `cos^-1sqrt3x+cos^-1x=pi/2`


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


If −1 < x < 0, then write the value of `sin^-1((2x)/(1+x^2))+cos^-1((1-x^2)/(1+x^2))`


Write the value of sin−1 \[\left( \cos\frac{\pi}{9} \right)\]


Write the value of cos−1 \[\left( \cos\frac{5\pi}{4} \right)\]


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


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


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


Write the value of  \[\tan^{- 1} \left( \frac{1}{x} \right)\]  for x < 0 in terms of `cot^-1x`


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

 

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


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


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×