English

Prove that tan^(-1)((6x-8x^3)/(1-12x^2))-tan^(-1)((4x)/(1-4x^2))=tan^(-1)2x;|2x|<1/√3 - Mathematics

Advertisements
Advertisements

Question

 
 
 

Prove that `tan^(-1)((6x-8x^3)/(1-12x^2))-tan^(-1)((4x)/(1-4x^2))=tan^(-1)2x;|2x|<1/sqrt3`

 
 
 
Advertisements

Solution

 
 
 

Consider the left hand side

L.H.S = `tan^(-1)((6x-8x^3)/(1-12x^2))-tan^(-1)((4x)/(1-4x^2))`

We know that, 

`tan^(-1)(A)-tan^(-1)(B)= tan^(-1)((A-B)/(1+AB))`

Thus, L.H.S = `tan^(-1)(((6x-8x^3)/(1-12x^2)-(4x)/(1-4x^2))/(1+((6x-8x^3)/(1-12x^2))((4x)/(1-4x^2))))`

`=tan^(-1)(((6x-8x^3)(1-4x^2)-4x(1-12x^2))/(((1-12x^2)(1-4x^2))/(1+(4x(6x-8x^3))/((1-12x^2)(1-4x^2)))))`

`=tan^(-1)((((6x-8x^3)(1-4x^2)-4x(1-12x^2))/((1-12x^2)(1-4x^2)))/(((1-12x^2)(1-4x^2)+4x(6x-8x^3))/((1-12x^2)(1-4x^2))))`

`=tan^(-1)(((6x-8x^3)(1-4x^2)-4x(1-12x^2))/((1-12x^2)(1-4x^2)+4x(6x-8x^3)))`

`=tan^(-1)((6x-24x^3-8x^3+32x^5-4x+48x^3)/(1-4x^2-12x^2+48x^4+24x^2-32x^4))`

 `=tan^(-1)((32x^5+16x^3+2x)/(16x^4+8x^2+1))`

 `=tan^(-1)((2x(16x^4+8x^2+1))/(16x^4+8x^2+1))`

 = tan-12x

 Thus, L.H.S=R.H.S

 
 
 
shaalaa.com
  Is there an error in this question or solution?
2015-2016 (March) All India Set 1 N

RELATED QUESTIONS

 

Prove that:

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

 

If `tan^-1(2x)+tan^-1(3x)=pi/4`, then find the value of ‘x’.


Prove the following:

3 sin−1 x = sin−1 (3x − 4x3), `x ∈ [-1/2, 1/2]`


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


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


Write the following function in the simplest form:

`tan^(-1)  (sqrt(1+x^2) -1)/x`, x ≠ 0


Write the function in the simplest form: `tan^(-1)  1/(sqrt(x^2 - 1)), |x| > 1`


Prove that:

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


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


sin–1 (1 – x) – 2 sin–1 x = `pi/2`, then x is equal to ______.


If y = `(x sin^-1 x)/sqrt(1 -x^2)`, prove that: `(1 - x^2)dy/dx = x + y/x`


Find: ∫ sin x · log cos x dx


Find the value, if it exists. If not, give the reason for non-existence

`sin^-1 [sin 5]`


Simplify: `tan^-1  x/y - tan^-1  (x - y)/(x + y)`


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


If a1, a2, a3,...,an is an arithmetic progression with common difference d, then evaluate the following expression.

`tan[tan^-1("d"/(1 + "a"_1 "a"_2)) + tan^-1("d"/(21 + "a"_2 "a"_3)) + tan^-1("d"/(1 + "a"_3 "a"_4)) + ... + tan^-1("d"/(1 + "a"_("n" - 1) "a""n"))]`


The maximum value of sinx + cosx is ____________.


If `"tan"^-1 ("cot"  theta) = 2theta, "then"  theta` is equal to ____________.


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


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


If tan-1 2x + tan-1 3x = `pi/4,` then x is ____________.


sin (tan−1 x), where |x| < 1, is equal to:


`"tan" (pi/4 + 1/2 "cos"^-1 "x") + "tan" (pi/4 - 1/2 "cos"^-1 "x") =` ____________.


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


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


`sin^-1(1 - x) - 2sin^-1 x = pi/2`, tan 'x' is equal to


Find the value of `sin^-1 [sin((13π)/7)]`


`tan(2tan^-1  1/5 + sec^-1  sqrt(5)/2 + 2tan^-1  1/8)` is equal to ______.


Solve for x: `sin^-1(x/2) + cos^-1x = π/6`


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×