Advertisements
Advertisements
प्रश्न
Prove that `tan^(-1)((6x-8x^3)/(1-12x^2))-tan^(-1)((4x)/(1-4x^2))=tan^(-1)2x;|2x|<1/sqrt3`
Advertisements
उत्तर
Consider the left hand side
L.H.S = `tan^(-1)((6x-8x^3)/(1-12x^2))-tan^(-1)((4x)/(1-4x^2))`
We know that,
`tan^(-1)(A)-tan^(-1)(B)= tan^(-1)((A-B)/(1+AB))`
Thus, L.H.S = `tan^(-1)(((6x-8x^3)/(1-12x^2)-(4x)/(1-4x^2))/(1+((6x-8x^3)/(1-12x^2))((4x)/(1-4x^2))))`
`=tan^(-1)(((6x-8x^3)(1-4x^2)-4x(1-12x^2))/(((1-12x^2)(1-4x^2))/(1+(4x(6x-8x^3))/((1-12x^2)(1-4x^2)))))`
`=tan^(-1)((((6x-8x^3)(1-4x^2)-4x(1-12x^2))/((1-12x^2)(1-4x^2)))/(((1-12x^2)(1-4x^2)+4x(6x-8x^3))/((1-12x^2)(1-4x^2))))`
`=tan^(-1)(((6x-8x^3)(1-4x^2)-4x(1-12x^2))/((1-12x^2)(1-4x^2)+4x(6x-8x^3)))`
`=tan^(-1)((6x-24x^3-8x^3+32x^5-4x+48x^3)/(1-4x^2-12x^2+48x^4+24x^2-32x^4))`
`=tan^(-1)((32x^5+16x^3+2x)/(16x^4+8x^2+1))`
`=tan^(-1)((2x(16x^4+8x^2+1))/(16x^4+8x^2+1))`
= tan-12x
Thus, L.H.S=R.H.S
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.
Write the following function in the simplest form:
`tan^(-1) (sqrt((1-cos x)/(1 + cos x)))`, 0 < x < π
if `tan^(-1) (x-1)/(x - 2) + tan^(-1) (x + 1)/(x + 2) = pi/4` then find the value of x.
Prove that:
`cos^(-1) 12/13 + sin^(-1) 3/5 = sin^(-1) 56/65`
Prove that:
`tan^(-1) sqrtx = 1/2 cos^(-1) (1-x)/(1+x)`, x ∈ [0, 1]
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))
Solve: `tan^-1x = cos^-1 (1 - "a"^2)/(1 + "a"^2) - cos^-1 (1 - "b"^2)/(1 + "b"^2), "a" > 0, "b" > 0`
Evaluate tan (tan–1(– 4)).
Prove that `2sin^-1 3/5 - tan^-1 17/31 = pi/4`
Prove that `sin^-1 8/17 + sin^-1 3/5 = sin^-1 7/85`
Show that `tan(1/2 sin^-1 3/4) = (4 - sqrt(7))/3` and justify why the other value `(4 + sqrt(7))/3` is ignored?
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 ______.
The value of cot–1(–x) for all x ∈ R in terms of cot–1x is ______.
The value of cos215° - cos230° + cos245° - cos260° + cos275° is ______.
The value of `"tan"^ -1 (3/4) + "tan"^-1 (1/7)` is ____________.
`"cot" (pi/4 - 2 "cot"^-1 3) =` ____________.
`"tan"^-1 1 + "cos"^-1 ((-1)/2) + "sin"^-1 ((-1)/2)`
The value of cot `("cosec"^-1 5/3 + "tan"^-1 2/3)` is ____________.
The value of sin (2tan-1 (0.75)) is equal to ____________.
The value of expression 2 `"sec"^-1 2 + "sin"^-1 (1/2)`
`"cot" ("cosec"^-1 5/3 + "tan"^-1 2/3) =` ____________.
`"tan"^-1 1/3 + "tan"^-1 1/5 + "tan"^-1 1/7 = "tan"^-1 1/8 =` ____________.
sin (tan−1 x), where |x| < 1, is equal to:
Simplest form of `tan^-1 ((sqrt(1 + cos "x") + sqrt(1 - cos "x"))/(sqrt(1 + cos "x") - sqrt(1 - cos "x")))`, `π < "x" < (3π)/2` is:
If `"sin" {"sin"^-1 (1/2) + "cos"^-1 "x"} = 1`, then the value of x is ____________.
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 ∠EAB = ________.
`sin^-1(1 - x) - 2sin^-1 x = pi/2`, tan 'x' is equal to
`tan(2tan^-1 1/5 + sec^-1 sqrt(5)/2 + 2tan^-1 1/8)` is equal to ______.
