Advertisements
Advertisements
प्रश्न
If cos–1α + cos–1β + cos–1γ = 3π, then α(β + γ) + β(γ + α) + γ(α + β) equals ______.
पर्याय
0
1
6
12
उत्तर
If cos–1α + cos–1β + cos–1γ = 3π, then α(β + γ) + β(γ + α) + γ(α + β) equals 6.
Explanation:
We have cos–1α + cos–1β + cos–1γ = 3π
⇒ cos–1α + cos–1β + cos–1γ = π + π + π
⇒ cos–1α = π, cos–1β = π and cos–1γ = π
⇒ α = cos π, β = cos π and γ = cos π
∴ α = – 1, β = – 1 and γ = – 1
Which gives α = β = γ = –1
So α(β + γ) + β(γ + α) + γ(α + β)
⇒ (– 1)(– 1 – 1) + (– 1)(– 1 – 1) + (– 1)(– 1 – 1)
⇒ (– 1)(– 2) + (– 1)(– 2) + (– 1)(– 2)
⇒ 2 + 2 + 2
⇒ 6
APPEARS IN
संबंधित प्रश्न
Prove that: `tan^(-1)(1/2)+tan^(-1)(1/5)+tan^(-1)(1/8)=pi/4`
Prove that `tan^(-1)((6x-8x^3)/(1-12x^2))-tan^(-1)((4x)/(1-4x^2))=tan^(-1)2x;|2x|<1/sqrt3`
Prove that:
`tan^(-1)""1/5+tan^(-1)""1/7+tan^(-1)""1/3+tan^(-1)""1/8=pi/4`
Write the following function in the simplest form:
`tan^(-1) (sqrt((1-cos x)/(1 + cos x))), x < pi`
Write the following function in the simplest form:
`tan^(-1) x/(sqrt(a^2 - x^2))`, |x| < a
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`
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`
Prove that `3sin^(-1)x = sin^(-1) (3x - 4x^3)`, `x in [-1/2, 1/2]`
If cos-1 x + cos -1 y + cos -1 z = π , prove that x2 + y2 + z2 + 2xyz = 1.
Solve: `cot^-1 x - cot^-1 (x + 2) = pi/12, x > 0`
Choose the correct alternative:
If `cot^-1(sqrt(sin alpha)) + tan^-1(sqrt(sin alpha))` = u, then cos 2u is equal to
Choose the correct alternative:
If |x| ≤ 1, then `2tan^-1x - sin^-1 (2x)/(1 + x^2)` is equal to
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 |x| ≤ 1, then `2 tan^-1x + sin^-1 ((2x)/(1 + x^2))` is equal to ______.
The value of `"tan"^-1 (1/2) + "tan"^-1 (1/3) + "tan"^-1 (7/8)` is ____________.
`"cot" ("cosec"^-1 5/3 + "tan"^-1 2/3) =` ____________.
If x = a sec θ, y = b tan θ, then `("d"^2"y")/("dx"^2)` at θ = `π/6` is:
If `"tan"^-1 2 "x + tan"^-1 3 "x" = pi/4`, then x is ____________.
`"tan"^-1 1/3 + "tan"^-1 1/5 + "tan"^-1 1/7 + "tan"^-1 1/8 =` ____________.
`"cos" (2 "tan"^-1 1/7) - "sin" (4 "sin"^-1 1/3) =` ____________.
`"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 ____________.
If `3 "sin"^-1 ((2"x")/(1 + "x"^2)) - 4 "cos"^-1 ((1 - "x"^2)/(1 + "x"^2)) + 2 "tan"^-1 ((2"x")/(1 - "x"^2)) = pi/3` then x is equal to ____________.
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 = ________.
What is the value of cos (sec–1x + cosec–1x), |x| ≥ 1
Write the following function in the simplest form:
`tan^-1 ((cos x - sin x)/(cos x + sin x)), (-pi)/4 < x < (3 pi)/4`
Solve:
sin–1(x) + sin–1(1 – x) = cos–1x.