English

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

Advertisements
Advertisements

Question

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

Solution

To prove:  `tan^(-1)  2/11 + tan^(-1)  7/24 = tan^(-1)  1/2`

L.H.S =` tan^(-1)  2/11 + tan^(-1)  7/24`

`= tan^(-1) (2/11 + 7/24)/(1-2/11. 7/24)`    `[tan^(-1) x + tan^(-1) y = tan^(-1)  (x + y)/(1 - xy)]`

= tan^(-1)  `((48+77)/(11xx24))/((11xx24 - 14)/(11xx24))`

`= tan^(-1)  (48 + 77)/(264 - 14) = tan^(-1)  125/250 = tan^(-1)  1/2 =`  R.H.S

shaalaa.com
  Is there an error in this question or solution?
Chapter 2: Inverse Trigonometric Functions - Exercise 2.2 [Page 47]

APPEARS IN

NCERT Mathematics [English] Class 12
Chapter 2 Inverse Trigonometric Functions
Exercise 2.2 | Q 3 | Page 47

RELATED QUESTIONS

Prove that `2tan^(-1)(1/5)+sec^(-1)((5sqrt2)/7)+2tan^(-1)(1/8)=pi/4`


Solve for x : tan-1 (x - 1) + tan-1x + tan-1 (x + 1) = tan-1 3x


Prove the following:

`3sin^(-1) x = sin^(-1)(3x - 4x^3), x in [-1/2, 1/2]`


Prove the following: 

`3cos^(-1) x = cos^(-1)(4x^3 - 3x), x in [1/2, 1]`


Write the following function in the simplest form:

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


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`


Find the value of 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`


`cos^(-1) (cos  (7pi)/6)` is equal to ______.


Prove that:

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


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


If cos-1 x + cos -1 y + cos -1 z = π , prove that x2 + y2 + z2 + 2xyz = 1.


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

`tan^-1(sin(- (5pi)/2))`


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

`sin^-1 [sin 5]`


Find the value of the expression in terms of x, with the help of a reference triangle

cos (tan–1 (3x – 1))


Find the value of  `tan(sin^-1  3/5 + cot^-1  3/2)`


Prove that `tan^-1x + tan^-1y + tan^-1z = tan^-1[(x + y + z - xyz)/(1 - xy - yz - zx)]`


Find the number of solutions of the equation `tan^-1 (x - 1) + tan^-1x + tan^-1(x + 1) = tan^-1(3x)`


Choose the correct alternative:

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


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


Prove that cot–17 + cot–18 + cot–118 = cot–13


Solve the equation `sin^-1 6x + sin^-1 6sqrt(3)x = - pi/2`


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


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`


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 ______.


`"cot" (pi/4 - 2  "cot"^-1  3) =` ____________.


The value of cot-1 9 + cosec-1 `(sqrt41/4)` is given by ____________.


The value of `"tan"^-1 (1/2) + "tan"^-1(1/3) + "tan"^-1(7/8)` is ____________.


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


`"tan"^-1 (sqrt3)`


Solve for x : `{"x cos" ("cot"^-1 "x") + "sin" ("cot"^-1 "x")}^2` = `51/50


The Government of India is planning to fix a hoarding board at the face of a building on the road of a busy market for awareness on COVID-19 protocol. Ram, Robert and Rahim are the three engineers who are working on this project. “A” is considered to be a person viewing the hoarding board 20 metres away from the building, standing at the edge of a pathway nearby. Ram, Robert and Rahim suggested to the firm to place the hoarding board at three different locations namely C, D and E. “C” is at the height of 10 metres from the ground level. For viewer A, the angle of elevation of “D” is double the angle of elevation of “C” The angle of elevation of “E” is triple the angle of elevation of “C” for the same viewer. Look at the figure given and based on the above information answer the following:

Measure of ∠CAB = ________.


The Government of India is planning to fix a hoarding board at the face of a building on the road of a busy market for awareness on COVID-19 protocol. Ram, Robert and Rahim are the three engineers who are working on this project. “A” is considered to be a person viewing the hoarding board 20 metres away from the building, standing at the edge of a pathway nearby. Ram, Robert and Rahim suggested to the firm to place the hoarding board at three different locations namely C, D and E. “C” is at the height of 10 metres from the ground level. For viewer A, the angle of elevation of “D” is double the angle of elevation of “C” The angle of elevation of “E” is triple the angle of elevation of “C” for the same viewer. Look at the figure given and based on the above information answer the following:

Measure of ∠DAB = ________.


The Government of India is planning to fix a hoarding board at the face of a building on the road of a busy market for awareness on COVID-19 protocol. Ram, Robert and Rahim are the three engineers who are working on this project. “A” is considered to be a person viewing the hoarding board 20 metres away from the building, standing at the edge of a pathway nearby. Ram, Robert and Rahim suggested to the firm to place the hoarding board at three different locations namely C, D and E. “C” is at the height of 10 metres from the ground level. For viewer A, the angle of elevation of “D” is double the angle of elevation of “C” The angle of elevation of “E” is triple the angle of elevation of “C” for the same viewer. Look at the figure given and based on the above information answer the following:

ЁЭР┤' Is another viewer standing on the same line of observation across the road. If the width of the road is 5 meters, then the difference between ∠CAB and ∠CA'B is ______.


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


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


Share
Notifications

Englishрд╣рд┐рдВрджреАрдорд░рд╛рдареА


      Forgot password?
Use app×