English

If | x | < 1, then prove that x2xx2xxxx2tan-1x=tan-1(2x1-x2)=sin-1(2x1+x2)=cos-1(1-x21+x2) - Mathematics and Statistics

Advertisements
Advertisements

Question

If | x | < 1, then prove that

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

Sum
Advertisements

Solution

Let tan-1x = y

Then, x = tan y

Now, `tan^-1 ("2x"/(1 - "x"^2)) = tan^-1  (("2 tan y")/(1 - tan^2 "y"))`

`= tan^-1 (tan 2"y")`

= 2y

= 2 tan-1x         ......(1)

`sin^-1 ("2x"/(1 + "x"^2)) = sin^-1 (("2 tan y")/(1 + tan^2 "y"))`

`= sin^-1 (sin 2"y")`

= 2y

= 2 tan-1x         ......(2)

`cos^-1 ((1 - "x"^2)/(1 + "x"^2)) = cos^-1 ((1 - tan^2 "y")/(1 + tan^2 "y"))`

`= cos^-1 (cos "2y")`

= 2y

`= 2 tan^-1 "x"`        ......(3)

From (1), (2) and (3), we get

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

shaalaa.com
Trigonometric Equations and Their Solutions
  Is there an error in this question or solution?
Chapter 3: Trigonometric Functions - Miscellaneous exercise 3 [Page 111]

RELATED QUESTIONS

Find the principal solution of the following equation: 

cosθ = `(1)/(2)`


Find the principal solution of the following equation: 

Sec θ = `(2)/sqrt(3)`


Find the principal solution of the following equation:

cot θ = 0


Find the principal solution of the following equation:

sin θ = `-1/2`


Find the principal solution of the following equation: 

tan θ = – 1


Find the general solution of the following equation:

sinθ = `1/2`.


Find the general solution of the following equation:

4 cos2 θ  = 3


Find the general solution of the following equation:

sin θ = tan θ


With the usual notations prove that `2{asin^2  "C"/(2) + "c"sin^2  "A"/(2)}` = a – b + c.


In Δ ABC, prove that a3 sin(B – C) + b3sin(C – A) + c3sin(A – B) = 0


With usual notations prove that 2(bc cosA + ac cosB + ab cosC) = a2 + b2 + c.


Select the correct option from the given alternatives:

The general solution of sec x = `sqrt(2)` is ______.


Select the correct option from the given alternatives:

If cos pθ = cos qθ, p ≠ q, then ______.


Select the correct option from the given alternatives:

In ΔABC, ac cos B - bc cos A = _______


`"cos"^-1 ("cos" (7pi)/6)` = _________.


The principal value of sin1 `(- sqrt3/2)` is ______.


If `"sin"^-1 4/5 + "cos"^-1 12/13 = "sin"^-1 alpha`, then α = ______.


`cos[tan^-1  1/3 + tan^-1  1/2]` = ______


If tan θ + tan 2θ + tan 3θ = tan θ.tan 2θ. tan 3θ, then the general value of the θ is ______.


Find the principal solutions of the following equation:

sin 2θ = `-1/2`


Find the general solutions of the following equation:

`tan theta = - sqrt3`


If sin-1(1 - x) - 2 sin-1x =  `pi/2`, then find the value of x.


Show that `tan^-1  1/2 - tan^-1  1/4 = tan^-1  2/9`.


Show that `cot^-1  1/3 - tan^-1  1/3 = cot^-1  3/4`.


Prove the following:

`cos^-1 "x" = tan^-1 (sqrt(1 - "x"^2)/"x")`, if x > 0


The principal solutions of `sqrt(3)` sec x − 2 = 0 are ______


Find the principal solutions of cosec x = 2


Find the principal solutions of cos 2ЁЭСе = 1


If sin-1 x = `pi/10`, for some x ∈ [-1, 1], then the value of cos-1 x is _______.


`int (sin (log x))^2/x` log x dx = ?


If f(x) = sin-1`(sqrt((1 - x)/2))`, then f'(x) = ?


The number of solutions of cos 2θ = sin θ in (0, 2π) is ______


If function

f(x) = `x - |x|/x, x < 0`
      = `x + |x|/x, x > 0`
      = 1, x = 0, then


The general solution of the equation tan θ tan 2θ = 1 is given by ______ 


The general solution of cosec x = `-sqrt2` is ______ 


The general solution of x(1 + y2)1/2 dx + y(1 + x2)1/2 dy = 0 is ______.


Which of the following equation has no solution?


The number of principal solutions of tan 2θ = 1 is ______.


Principal solutions at the equation sin 2x + cos 2x = 0, where π < x < 2 π are ______.


The general solution to cos100x – sin100x = 1 is ______.


If `2sin^-1  3/7` = cos–1β, then find the value of β.


If tan θ + sec θ = `sqrt(3)`, find the general value of θ.


Share
Notifications

Englishрд╣рд┐рдВрджреАрдорд░рд╛рдареА


      Forgot password?
Use app×