Advertisements
Advertisements
प्रश्न
Find the number of solutions of the equation `tan^-1 (x - 1) + tan^-1x + tan^-1(x + 1) = tan^-1(3x)`
Advertisements
उत्तर
tan–1(x – 1) + tan–1 x + tan–1(x + 1)
= tan–1(x – 1) + tan–1(x + 1) + tan–1x
= `tan^-1 [(x - 1 + x + 1)/(1 - (x - 1)(x + 1))] + tan^-1x`
= `tan^-1 [(2x)/(1 - (x^2 - 1))] + tan^-1x`
= `tan^-1 [(2x)/(1 - x^2 + 1)] + tan^-1x`
= `tan^-1 [(2x)/(2 - x^2)] + tan^-1x`
= `tan^-1 [((2x)/(2 - x^2) + x)/(1 - (2x)/(2 - x^2) * x)]`
= `tan^-1 [((2x + 2x - x^3)/(2 - x^2))/((2 - x^2 - 2x^2)/(2 - x^2))]`
= `tan^-1 [(4x - x^3)/(2 - 3x^2)]`
Given L.H.S. = R.H.S
`tan^-1 [(4x - x^3)/(2 - 3x^2)] = tan^-1 3x`
`(4x - x^3)/(2 - 3x)` = 3x
4x – x3 = 6x – 9x3
8x3 = 2x
8x3 – 2x = 0
2x(x2 – 1) = 0
x = 0, x2 = 1
x = ±1
Number of solutions are three (0, 1 – 1)
APPEARS IN
संबंधित प्रश्न
if `sin(sin^(-1) 1/5 + cos^(-1) x) = 1` then find the value of x
Find the value of the given expression.
`tan^(-1) (tan (3pi)/4)`
Solve `tan^(-1) - tan^(-1) (x - y)/(x+y)` is equal to
(A) `pi/2`
(B). `pi/3`
(C) `pi/4`
(D) `(-3pi)/4`
Solve for x : \[\tan^{- 1} \left( \frac{x - 2}{x - 1} \right) + \tan^{- 1} \left( \frac{x + 2}{x + 1} \right) = \frac{\pi}{4}\] .
Prove that
\[2 \tan^{- 1} \left( \frac{1}{5} \right) + \sec^{- 1} \left( \frac{5\sqrt{2}}{7} \right) + 2 \tan^{- 1} \left( \frac{1}{8} \right) = \frac{\pi}{4}\] .
Find the value of the expression in terms of x, with the help of a reference triangle
sin (cos–1(1 – x))
Prove that `sin^-1 3/5 - cos^-1 12/13 = sin^-1 16/65`
Solve: `tan^-1x = cos^-1 (1 - "a"^2)/(1 + "a"^2) - cos^-1 (1 - "b"^2)/(1 + "b"^2), "a" > 0, "b" > 0`
Choose the correct alternative:
If |x| ≤ 1, then `2tan^-1x - sin^-1 (2x)/(1 + x^2)` is equal to
Evaluate: `tan^-1 sqrt(3) - sec^-1(-2)`.
Evaluate: `sin^-1 [cos(sin^-1 sqrt(3)/2)]`
If |x| ≤ 1, then `2 tan^-1x + sin^-1 ((2x)/(1 + x^2))` is equal to ______.
If cos–1α + cos–1β + cos–1γ = 3π, then α(β + γ) + β(γ + α) + γ(α + β) equals ______.
If `"tan"^-1 (("x" - 1)/("x" + 2)) + "tan"^-1 (("x" + 1)/("x" + 2)) = pi/4,` then x is equal to ____________.
`"tan" (pi/4 + 1/2 "cos"^-1 "x") + "tan" (pi/4 - 1/2 "cos"^-1 "x") =` ____________.
`"sin"^-1 (1 - "x") - 2 "sin"^-1 "x" = pi/2`
If `"sin" {"sin"^-1 (1/2) + "cos"^-1 "x"} = 1`, then the value of x is ____________.
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:
𝐴' 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 ______.
Find the value of `tan^-1 [2 cos (2 sin^-1 1/2)] + tan^-1 1`.
If \[\tan^{-1}\left(\frac{x}{2}\right)+\tan^{-1}\left(\frac{y}{2}\right)+\tan^{-1}\left(\frac{z}{2}\right)=\frac{\pi}{2}\] then xy + yz + zx =
