हिंदी

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
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 4: Inverse Trigonometric Functions - Exercise 4.11 [पृष्ठ ८२]

APPEARS IN

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

वीडियो ट्यूटोरियलVIEW ALL [2]

संबंधित प्रश्न

Find the value of the following: `tan(1/2)[sin^(-1)((2x)/(1+x^2))+cos^(-1)((1-y^2)/(1+y^2))],|x| <1,y>0 and xy <1`


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


Solve for x:

`2tan^(-1)(cosx)=tan^(-1)(2"cosec" x)`


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


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(sin  (7pi)/6)`


`sin^-1(sin4)`


Evaluate the following:

`tan^-1(tan4)`


Evaluate the following:

`cosec^-1(cosec  (13pi)/6)`


Write the following in the simplest form:

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


Write the following in the simplest form:

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


Write the following in the simplest form:

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


Evaluate the following:

`sin(sin^-1  7/25)`

 


Evaluate:

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


Evaluate:

`cos(tan^-1  3/4)`


Evaluate:

`sin(tan^-1x+tan^-1  1/x)` for x > 0


Evaluate:

`cos(sec^-1x+\text(cosec)^-1x)`,|x|≥1


`sin(sin^-1  1/5+cos^-1x)=1`


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


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


Find the value of `tan^-1  (x/y)-tan^-1((x-y)/(x+y))`


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


Prove that: `cos^-1  4/5+cos^-1  12/13=cos^-1  33/65`


`tan^-1  1/7+2tan^-1  1/3=pi/4`


Prove that

`sin{tan^-1  (1-x^2)/(2x)+cos^-1  (1-x^2)/(2x)}=1`


Find the value of the following:

`tan^-1{2cos(2sin^-1  1/2)}`


Solve the following equation for x:

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


Write the value of `sin^-1((-sqrt3)/2)+cos^-1((-1)/2)`


Write the value of cos−1 (cos 1540°).


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


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


Write the principal value of `sin^-1(-1/2)`


Write the value of \[\cos^{- 1} \left( \cos\frac{14\pi}{3} \right)\]


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


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


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

 

The value of \[\cos^{- 1} \left( \cos\frac{5\pi}{3} \right) + \sin^{- 1} \left( \sin\frac{5\pi}{3} \right)\] is

 


In a ∆ ABC, if C is a right angle, then
\[\tan^{- 1} \left( \frac{a}{b + c} \right) + \tan^{- 1} \left( \frac{b}{c + a} \right) =\]

 

 


The value of sin \[\left( \frac{1}{4} \sin^{- 1} \frac{\sqrt{63}}{8} \right)\] is

 


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×