Advertisements
Advertisements
Question
Prove that:
`cot^(-1) ((sqrt(1+sin x) + sqrt(1-sinx))/(sqrt(1+sin x) - sqrt(1- sinx))) = x/2, x in (0, pi/4)`
Advertisements
Solution
L.H.S. = `cot^(-1) ((sqrt(1+sin x) + sqrt(1-sinx))/(sqrt(1+sin x) - sqrt(1- sinx)))`
= `cot^(-1) (sqrt(1+sin x) + sqrt(1-sinx))/(sqrt(1+sin x) - sqrt(1 - sin x)) xx (sqrt(1+sin x) + sqrt(1-sinx))/(sqrt(1+sin x) - sqrt(1 - sin x))`
= `cot^(-1) ((1+sinx) + (1-sinx) + 2sqrt(1 - sin^2 x))/((1+sinx) - (1 - sinx)`
= `cot^(-1) (2(1 + cos x))/(2sin x)`
= `cot^(-1) (1+ cosx)/sin x`
= `cot^(-1) (2 cos^2 x/2)/(2sin x/2 cos x/2)`
= `cot^-1 (cot x/2)`
= `x/2`
L.H.S. = R.H.S.
APPEARS IN
RELATED QUESTIONS
Prove that:
`tan^(-1)""1/5+tan^(-1)""1/7+tan^(-1)""1/3+tan^(-1)""1/8=pi/4`
Prove the following:
3cos−1x = cos−1(4x3 − 3x), `x ∈ [1/2, 1]`
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 `cot(tan^(-1) a + cot^(-1) a)`
Find the value of the 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
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]
sin (tan–1 x), |x| < 1 is equal to ______.
Solve the following equation for x: `cos (tan^(-1) x) = sin (cot^(-1) 3/4)`
If cos-1 x + cos -1 y + cos -1 z = π , prove that x2 + y2 + z2 + 2xyz = 1.
If y = `(x sin^-1 x)/sqrt(1 -x^2)`, prove that: `(1 - x^2)dy/dx = x + y/x`
Prove that `tan^-1x + tan^-1 (2x)/(1 - x^2) = tan^-1 (3x - x^3)/(1 - 3x^2), |x| < 1/sqrt(3)`
Choose the correct alternative:
`sin^-1 3/5 - cos^-1 13/13 + sec^-1 5/3 - "cosec"^-1 13/12` is equal to
Evaluate tan (tan–1(– 4)).
If `tan^-1x = pi/10` for some x ∈ R, then the value of cot–1x is ______.
If 3 tan–1x + cot–1x = π, then x equals ______.
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) =` ____________.
The value of expression 2 `"sec"^-1 2 + "sin"^-1 (1/2)`
The value of the expression tan `(1/2 "cos"^-1 2/sqrt3)`
`"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 =` ____________.
If `"tan"^-1 (("x" - 1)/("x" + 2)) + "tan"^-1 (("x" + 1)/("x" + 2)) = pi/4,` then x is equal to ____________.
The value of cot-1 9 + cosec-1 `(sqrt41/4)` is given by ____________.
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 ____________.
`"cos"^-1["cos"(2"cot"^-1(sqrt2 - 1))]` = ____________.
The value of `"cos"^-1 ("cos" ((33pi)/5))` is ____________.
`"cos"^-1 1/2 + 2 "sin"^-1 1/2` 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:
Domain and Range of tan-1 x = ________.
`sin^-1(1 - x) - 2sin^-1 x = pi/2`, tan 'x' is equal to
Find the value of `sin^-1 [sin((13π)/7)]`
Write the following function in the simplest form:
`tan^-1 ((cos x - sin x)/(cos x + sin x)), (-pi)/4 < x < (3 pi)/4`
`tan^-1 sqrt3 - cot^-1 (- sqrt3)` is equal to ______.
The value of cosec `[sin^-1((-1)/2)] - sec[cos^-1((-1)/2)]` is equal to ______.
If sin–1x + sin–1y + sin–1z = π, show that `x^2 - y^2 - z^2 + 2yzsqrt(1 - x^2) = 0`
