Advertisements
Advertisements
प्रश्न
Prove that `cot^(-1) ((sqrt(1 + sin x) + sqrt(1 - sinx))/(sqrt(1 + sin x) - sqrt(1 - sinx))) = x/2, x ∈ (0, pi/4)`.
Advertisements
उत्तर
Consider `(sqrt(1 + sinx) + sqrt(1 - sin x))/(sqrt(1 + sinx) - sqrt(1 - sinx))`
= `((sqrt(1 + sinx) + sqrt(1 - sinx))^2)/((sqrt(1 + sin x))^2 - (sqrt(1 - sin x))^2)` ...(By rationalizing)
= `((1 + sinx) + (1 - sinx) + 2sqrt((1 + sinx)(1 - sinx)))/(1 + sinx - 1 + sinx)`
= `(2(1 + sqrt(1 - sin^2x)))/(2sinx)`
= `(1 + cosx)/(sin x)`
= `(2 cos^2 x/2)/(2sin x/2 cos x/2)`
= `cot x/2`
∴ L.H.S = `cot^(-1) ((sqrt(1 + sin x) + sqrt(1 - sinx))/(sqrt(1 + sin x) - sqrt(1 - sinx)))`
= `cot^(-1) (cot x/2)`
= `x/2` = R.H.S.
APPEARS IN
संबंधित प्रश्न
If `sin (sin^(−1)(1/5)+cos^(−1) x)=1`, then find the value of x.
Prove that `cot^(-1)((sqrt(1+sinx)+sqrt(1-sinx))/(sqrt(1+sinx)-sqrt(1-sinx)))=x/2;x in (0,pi/4) `
Solve for x : tan-1 (x - 1) + tan-1x + tan-1 (x + 1) = tan-1 3x
Prove that `tan^(-1)((6x-8x^3)/(1-12x^2))-tan^(-1)((4x)/(1-4x^2))=tan^(-1)2x;|2x|<1/sqrt3`
If `tan^-1(2x)+tan^-1(3x)=pi/4`, then find the value of ‘x’.
Write the following function in the simplest form:
`tan^(-1) x/(sqrt(a^2 - x^2)), |x| < 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`
sin (tan–1 x), |x| < 1 is equal to ______.
Solve the following equation for x: `cos (tan^(-1) x) = sin (cot^(-1) 3/4)`
Prove that `3sin^(-1)x = sin^(-1) (3x - 4x^3)`, `x in [-1/2, 1/2]`
Solve for x : \[\tan^{- 1} \left( \frac{x - 2}{x - 1} \right) + \tan^{- 1} \left( \frac{x + 2}{x + 1} \right) = \frac{\pi}{4}\] .
Solve: tan-1 4 x + tan-1 6x `= π/(4)`.
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
Choose the correct alternative:
`tan^-1 (1/4) + tan^-1 (2/9)` is equal to
Choose the correct alternative:
sin–1(2 cos2x – 1) + cos–1(1 – 2 sin2x) =
Evaluate: `tan^-1 sqrt(3) - sec^-1(-2)`.
Evaluate: `sin^-1 [cos(sin^-1 sqrt(3)/2)]`
Evaluate `cos[sin^-1 1/4 + sec^-1 4/3]`
Prove that `2sin^-1 3/5 - tan^-1 17/31 = pi/4`
If a1, a2, a3,...,an is an arithmetic progression with common difference d, then evaluate the following expression.
`tan[tan^-1("d"/(1 + "a"_1 "a"_2)) + tan^-1("d"/(21 + "a"_2 "a"_3)) + tan^-1("d"/(1 + "a"_3 "a"_4)) + ... + tan^-1("d"/(1 + "a"_("n" - 1) "a""n"))]`
If cos–1α + cos–1β + cos–1γ = 3π, then α(β + γ) + β(γ + α) + γ(α + β) equals ______.
Solve for x : `"sin"^-1 2 "x" + sin^-1 3"x" = pi/3`
The value of `"tan"^ -1 (3/4) + "tan"^-1 (1/7)` is ____________.
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:
Solve for x : `"sin"^-1 2"x" + "sin"^-1 3"x" = pi/3`
The value of `"tan"^-1 (3/4) + "tan"^-1 (1/7)` is ____________.
`"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 `6"sin"^-1 ("x"^2 - 6"x" + 8.5) = pi,` 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 ∠EAB = ________.
Solve for x: `sin^-1(x/2) + cos^-1x = π/6`
If sin–1x + sin–1y + sin–1z = π, show that `x^2 - y^2 - z^2 + 2yzsqrt(1 - x^2) = 0`
Solve:
sin–1 (x) + sin–1 (1 – x) = cos–1 x
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 =
`sin (tan^-1 4/5 + tan^-1 4/3 + tan^-1 1/9 - tan^-1 1/7)` is equal to ______.
