Advertisements
Advertisements
प्रश्न
Prove that `2sin^-1 3/5 - tan^-1 17/31 = pi/4`
उत्तर
Let `sin^-1 3/5` = θ
Then sin θ = `3/5`
Where θ ∈ `[(-pi)/2, pi/2]`
Thus tan θ = `3/4`
Which gives θ = `tan^-1 3/4`.
Therefore, `2sin^-1 3/5 - tan^-1 17/31`
= `2theta - tan^-1 17/31`
= `2tan^-1 3/4 - tan^-1 17/31`
= `tan^-1 ((2 * 3/4)/(1 - 9/16)) - tan^-1 17/31`
= `tan^-1 24/7 - tan^-1 17/31`
= `tan^-1 ((24/7 - 17/31)/(1 + 24/7 * 17/31))`
= `pi/4`
APPEARS IN
संबंधित प्रश्न
If a line makes angles 90°, 60° and θ with x, y and z-axis respectively, where θ is acute, then find θ.
Prove the following:
`3cos^(-1) x = cos^(-1)(4x^3 - 3x), x in [1/2, 1]`
Find the value of the following:
`tan^-1 [2 cos (2 sin^-1 1/2)]`
Solve `tan^(-1) - tan^(-1) (x - y)/(x+y)` is equal to
(A) `pi/2`
(B). `pi/3`
(C) `pi/4`
(D) `(-3pi)/4`
Find the value of the expression in terms of x, with the help of a reference triangle
cos (tan–1 (3x – 1))
Prove that `tan^-1 2/11 + tan^-1 7/24 = tan^-1 1/2`
Prove that `tan^-1x + tan^-1y + tan^-1z = tan^-1[(x + y + z - xyz)/(1 - xy - yz - zx)]`
If tan–1x + tan–1y + tan–1z = π, show that x + y + z = xyz
Prove that `tan^-1x + tan^-1 (2x)/(1 - x^2) = tan^-1 (3x - x^3)/(1 - 3x^2), |x| < 1/sqrt(3)`
Solve: `cot^-1 x - cot^-1 (x + 2) = pi/12, x > 0`
Choose the correct alternative:
If |x| ≤ 1, then `2tan^-1x - sin^-1 (2x)/(1 + x^2)` is equal to
Choose the correct alternative:
If `sin^-1x + cot^-1 (1/2) = pi/2`, then x is equal to
Choose the correct alternative:
sin(tan–1x), |x| < 1 is equal to
Solve the equation `sin^-1 6x + sin^-1 6sqrt(3)x = - pi/2`
If `sin^-1 ((2"a")/(1 + "a"^2)) + cos^-1 ((1 - "a"^2)/(1 + "a"^2)) = tan^-1 ((2x)/(1 - x^2))`. where a, x ∈ ] 0, 1, then the value of x is ______.
If |x| ≤ 1, then `2 tan^-1x + sin^-1 ((2x)/(1 + x^2))` is equal to ______.
If `"cot"^-1 (sqrt"cos" alpha) - "tan"^-1 (sqrt"cos" alpha) = "x",` the sinx is equal to ____________.
`"cos" (2 "tan"^-1 1/7) - "sin" (4 "sin"^-1 1/3) =` ____________.
`"tan"^-1 1/3 + "tan"^-1 1/5 + "tan"^-1 1/7 = "tan"^-1 1/8 =` ____________.
If tan-1 2x + tan-1 3x = `pi/4,` then x is ____________.
Solve for x : `"sin"^-1 2"x" + "sin"^-1 3"x" = pi/3`
`"tan" (pi/4 + 1/2 "cos"^-1 "x") + "tan" (pi/4 - 1/2 "cos"^-1 "x") =` ____________.
`"tan"^-1 (sqrt3)`
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:
Domain and Range of tan-1 x = ________.
What is the value of cos (sec–1x + cosec–1x), |x| ≥ 1
Find the value of `sin^-1 [sin((13π)/7)]`
If `tan^-1 ((x - 1)/(x + 1)) + tan^-1 ((2x - 1)/(2x + 1)) = tan^-1 (23/36)` = then prove that 24x2 – 23x – 12 = 0
