English

Prove that tan-1(1+x2+1-x2(1+x2)-1-x2)=π2+12cos-1x2 - Mathematics

Advertisements
Advertisements

Question

Prove that `tan^-1 ((sqrt(1 + x^2) + sqrt(1 - x^2))/((1 + x^2) - sqrt(1 - x^2))) = pi/2 + 1/2 cos^-1x^2`

Sum
Advertisements

Solution

L.H.S. `tan^-1 [(sqrt(1 + x^2) + sqrt(1 - x^2))/(sqrt(1 + x^2 - sqrt(1 - x^2)))]`

Put x2 = cos θ

∴ θ = `cos^-1 x^2`

⇒ `tan^-1 [(sqrt(1 + cos theta) + sqrt(1 - cos theta))/(sqrt(1 + cos theta) - sqrt(1 - cos theta))]`

⇒ `tan^-1 [sqrt(2cos^2  theta/2 + sqrt(2sin^2  theta/2))/(sqrt(2cos^2  theta/2 - sqrt(2sin^2   theta/2)))]`  ......`[(because 1 + cos theta = 2 cos^2  theta/2),(1 - cos theta = 2 sin^2  theta/2)]`

⇒ `tan^-1 [(cos  theta/2 + sin  theta/2),(cos  theta/2 - sin  theta/2)]`

⇒ `tan^-1 [(1 + tan  theta/2),(1 - tan  theta/2)]`  ......[Dividing the Nr. and Den. by cos θ/2]

⇒ `tan [tan(pi/4  theta/2)]`  ......`[because (1 + tan theta)/(1 - tan theta) = tan(pi/4 + theta)]`

⇒ `pi/4 + theta/2`

⇒ `pi/4 + 1/2 cos^-1 x^2` R.H.S. ......[Putting θ = cos–1x2]

Hence proved.

shaalaa.com
  Is there an error in this question or solution?
Chapter 2: Inverse Trigonometric Functions - Exercise [Page 36]

APPEARS IN

NCERT Exemplar Mathematics [English] Class 12
Chapter 2 Inverse Trigonometric Functions
Exercise | Q 12 | Page 36

RELATED QUESTIONS

Prove that `cot^(-1)((sqrt(1+sinx)+sqrt(1-sinx))/(sqrt(1+sinx)-sqrt(1-sinx)))=x/2;x in (0,pi/4) `


 

Prove that:

`tan^(-1)""1/5+tan^(-1)""1/7+tan^(-1)""1/3+tan^(-1)""1/8=pi/4`

 

Write the following function in the simplest form:

`tan^(-1) (sqrt((1-cos x)/(1 + cos x)))`, 0 < x < π


Write the following function in the simplest form:

`tan^(-1)  x/(sqrt(a^2 - x^2))`, |x| < a


Write the following function in the simplest form:

`tan^(-1) ((3a^2 x - x^3)/(a^3 - 3ax^2)), a > 0; (-a)/sqrt3 < x < a/sqrt3`


Prove that:

`cos^(-1)  4/5 + cos^(-1)  12/13 = cos^(-1)  33/65`


Solve the following equation:

2 tan−1 (cos x) = tan−1 (2 cosec x)


sin (tan–1 x), |x| < 1 is equal to ______.


Solve for x : \[\tan^{- 1} \left( \frac{x - 2}{x - 1} \right) + \tan^{- 1} \left( \frac{x + 2}{x + 1} \right) = \frac{\pi}{4}\] .


Solve: tan-1 4 x + tan-1 6x `= π/(4)`.


Prove that `tan^-1  2/11 + tan^-1  7/24 = tan^-1  1/2`


Choose the correct alternative:

If `sin^-1x + sin^-1y = (2pi)/3` ; then `cos^-1x + cos^-1y` is equal to


Choose the correct alternative:

If |x| ≤ 1, then `2tan^-1x - sin^-1  (2x)/(1 + x^2)` is equal to


Evaluate: `sin^-1 [cos(sin^-1 sqrt(3)/2)]`


Evaluate `cos[sin^-1  1/4 + sec^-1  4/3]`


Show that `2tan^-1 {tan  alpha/2 * tan(pi/4 - beta/2)} = tan^-1  (sin alpha cos beta)/(cosalpha + sinbeta)`


Evaluate `cos[cos^-1 ((-sqrt(3))/2) + pi/6]`


Show that `tan(1/2 sin^-1  3/4) = (4 - sqrt(7))/3` and justify why the other value `(4 + sqrt(7))/3` is ignored?


If y = `2 tan^-1x + sin^-1 ((2x)/(1 + x^2))` for all x, then ______ < y < ______.


The value of cot–1(–x) for all x ∈ R in terms of cot–1x is ______.


Solve for x : `"sin"^-1  2 "x" + sin^-1  3"x" = pi/3`


`"sin" {2  "cos"^-1 ((-3)/5)}` is equal to ____________.


If x = a sec θ, y = b tan θ, then `("d"^2"y")/("dx"^2)` at θ = `π/6` is:


Simplest form of `tan^-1 ((sqrt(1 + cos "x") + sqrt(1 - cos "x"))/(sqrt(1 + cos "x") - sqrt(1 - cos "x")))`, `π < "x" < (3π)/2` is:


`"cos" (2  "tan"^-1 1/7) - "sin" (4  "sin"^-1 1/3) =` ____________.


If `6"sin"^-1 ("x"^2 - 6"x" + 8.5) = pi,` then x is equal to ____________.


`"cos"^-1 1/2 + 2  "sin"^-1 1/2` is equal to ____________.


`tan^-1  1/2 + tan^-1  2/11` is equal to


The value of `tan^-1 (x/y) - tan^-1  (x - y)/(x + y)` is equal to


`50tan(3tan^-1(1/2) + 2cos^-1(1/sqrt(5))) + 4sqrt(2) tan(1/2tan^-1(2sqrt(2)))` is equal to ______.


The set of all values of k for which (tan–1 x)3 + (cot–1 x)3 = kπ3, x ∈ R, is the internal ______.


If `tan^-1 ((x - 1)/(x + 1)) + tan^-1 ((2x - 1)/(2x + 1)) = tan^-1 (23/36)` = then prove that 24x2 – 23x – 12 = 0


`tan^-1 sqrt3 - cot^-1 (- sqrt3)` is equal to ______.


Solve:

sin–1 (x) + sin–1 (1 – x) = cos–1 x


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×