Advertisements
Advertisements
प्रश्न
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?
Advertisements
उत्तर
We have `tan(1/2 sin^-1 3/4)`
Let `1/2 sin^-1 3/4` = θ
⇒ `sin^-1 3/4` = 2θ
⇒ sin 2θ = `3/4`
⇒ `(2 tan theta)/(1 + tan^2theta) = 3/4`
⇒ `3 tan theta^2 - 8` and θ ++ 3 = 0
⇒ tan θ = `(8 +- sqrt(64 - 36))/6`
⇒ tan θ = `(8 +- sqrt(28))/6 = (8 +- sqrt(7))/6 = (4 + sqrt(7))/3`
Now `- pi/2 ≤ sin^-1 3/4 ≤ pi/2`
⇒ `(-pi)/2 ≤ 1/2 sin^-1 3/4 ≤ pi/2`
∴ `tan((-pi)/2) ≤ tan(1/2(sin^-1 3/4)) ≤ tan pi/4`
⇒ `-1 ≤ tan (1/2 sin^-1 3/4) ≤ 1`
⇒ tan θ = `(4 - sqrt(7))/3` ....`(tan theta = (4 + sqrt(7))/3 > 1, "which is not possible")`
APPEARS IN
संबंधित प्रश्न
Solve for x : tan-1 (x - 1) + tan-1x + tan-1 (x + 1) = tan-1 3x
If `tan^-1(2x)+tan^-1(3x)=pi/4`, then find the value of ‘x’.
Prove `tan^(-1) 2/11 + tan^(-1) 7/24 = tan^(-1) 1/2`
Write the function in the simplest form: `tan^(-1) 1/(sqrt(x^2 - 1)), |x| > 1`
Write the function in the simplest form: `tan^(-1) ((cos x - sin x)/(cos x + sin x)) `,` 0 < x < pi`
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 the following:
`tan^-1 [2 cos (2 sin^-1 1/2)]`
if `sin(sin^(-1) 1/5 + cos^(-1) x) = 1` then find the value of x
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)`
sin–1 (1 – x) – 2 sin–1 x = `pi/2`, then x is equal to ______.
Prove that `tan {pi/4 + 1/2 cos^(-1) a/b} + tan {pi/4 - 1/2 cos^(-1) a/b} = (2b)/a`
Prove that `3sin^(-1)x = sin^(-1) (3x - 4x^3)`, `x in [-1/2, 1/2]`
Solve for x : \[\cos \left( \tan^{- 1} x \right) = \sin \left( \cot^{- 1} \frac{3}{4} \right)\] .
Find the value, if it exists. If not, give the reason for non-existence
`sin^-1 [sin 5]`
Find the value of `sin^-1[cos(sin^-1 (sqrt(3)/2))]`
Find the value of `cot[sin^-1 3/5 + sin^-1 4/5]`
Find the value of `tan(sin^-1 3/5 + cot^-1 3/2)`
Prove that `tan^-1x + tan^-1y + tan^-1z = tan^-1[(x + y + z - xyz)/(1 - xy - yz - zx)]`
Solve: `cot^-1 x - cot^-1 (x + 2) = pi/12, x > 0`
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) =
Choose the correct alternative:
sin(tan–1x), |x| < 1 is equal to
Evaluate tan (tan–1(– 4)).
Evaluate `cos[sin^-1 1/4 + sec^-1 4/3]`
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"))]`
The domain of the function defind by f(x) `= "sin"^-1 sqrt("x" - 1)` is ____________.
If tan-1 2x + tan-1 3x = `pi/4,` then x is ____________.
sin (tan−1 x), where |x| < 1, is equal to:
If `"tan"^-1 2 "x + tan"^-1 3 "x" = pi/4`, then x is ____________.
`"tan" (pi/4 + 1/2 "cos"^-1 "x") + "tan" (pi/4 - 1/2 "cos"^-1 "x") =` ____________.
`"tan"^-1 1/3 + "tan"^-1 1/5 + "tan"^-1 1/7 + "tan"^-1 1/8 =` ____________.
`"cos"^-1 1/2 + 2 "sin"^-1 1/2` is equal to ____________.
What is the simplest form of `tan^-1 sqrt(1 - x^2 - 1)/x, x ≠ 0`
Find the value of `sin^-1 [sin((13π)/7)]`
Find the value of `tan^-1 [2 cos (2 sin^-1 1/2)] + tan^-1 1`.
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 =
