Advertisements
Advertisements
प्रश्न
Find the value of x, if tan `[sec^(-1) (1/x) ] = sin ( tan^(-1) 2) , x > 0 `.
Advertisements
उत्तर
Let sec-1 `(1/x) = theta`
` ⇒ sec theta = 1/x`
⇒ cos θ = x
⇒ tan ` (sec^(-1) (1/x)) = tan theta = sqrt(1 -x^2 ) /x ` ...(1)

Now consider,
sin ( tan -1 2 )
Let tan-1 2 = Φ
tan Φ = 2
sin ( tan-1 2) = sin Φ = `2/sqrt(5) ` ...(ii)

From (i) and (ii)
`sqrt(1- x^2 )/x = 2/sqrt(5)`
5(1 - x2 ) = 4x2
`x = +- sqrt(5)/3 " but " x > 0 ⇒ x = sqrt(5)/3`
APPEARS IN
संबंधित प्रश्न
Solve the following for x :
`tan^(-1)((x-2)/(x-3))+tan^(-1)((x+2)/(x+3))=pi/4,|x|<1`
Find the domain of `f(x) =2cos^-1 2x+sin^-1x.`
Find the domain of `f(x)=cos^-1x+cosx.`
`sin^-1(sin2)`
Evaluate the following:
`cos^-1(cos4)`
Evaluate the following:
`tan^-1(tan pi/3)`
Evaluate the following:
`tan^-1(tan1)`
Evaluate the following:
`sec^-1(sec (25pi)/6)`
Evaluate the following:
`cosec^-1(cosec (6pi)/5)`
Evaluate the following:
`cot^-1(cot pi/3)`
Write the following in the simplest form:
`tan^-1(x/(a+sqrt(a^2-x^2))),-a<x<a`
Write the following in the simplest form:
`sin^-1{(x+sqrt(1-x^2))/sqrt2},-1<x<1`
Evaluate the following:
`sin(tan^-1 24/7)`
Evaluate the following:
`sec(sin^-1 12/13)`
Solve: `cos(sin^-1x)=1/6`
`2tan^-1 3/4-tan^-1 17/31=pi/4`
Prove that
`tan^-1((1-x^2)/(2x))+cot^-1((1-x^2)/(2x))=pi/2`
For any a, b, x, y > 0, prove that:
`2/3tan^-1((3ab^2-a^3)/(b^3-3a^2b))+2/3tan^-1((3xy^2-x^3)/(y^3-3x^2y))=tan^-1 (2alphabeta)/(alpha^2-beta^2)`
`where alpha =-ax+by, beta=bx+ay`
Write the value of tan−1 x + tan−1 `(1/x)` for x < 0.
Write the value of sin (cot−1 x).
Write the value of sin−1 (sin 1550°).
Write the value of cos−1 \[\left( \tan\frac{3\pi}{4} \right)\]
Write the value of \[\cos^{- 1} \left( \cos\frac{14\pi}{3} \right)\]
Write the value of `cot^-1(-x)` for all `x in R` in terms of `cot^-1(x)`
The positive integral solution of the equation
\[\tan^{- 1} x + \cos^{- 1} \frac{y}{\sqrt{1 + y^2}} = \sin^{- 1} \frac{3}{\sqrt{10}}\text{ is }\]
If α = \[\tan^{- 1} \left( \tan\frac{5\pi}{4} \right) \text{ and }\beta = \tan^{- 1} \left( - \tan\frac{2\pi}{3} \right)\] , then
\[\text{ If }\cos^{- 1} \frac{x}{3} + \cos^{- 1} \frac{y}{2} = \frac{\theta}{2}, \text{ then }4 x^2 - 12xy \cos\frac{\theta}{2} + 9 y^2 =\]
If \[\tan^{- 1} \left( \frac{1}{1 + 1 . 2} \right) + \tan^{- 1} \left( \frac{1}{1 + 2 . 3} \right) + . . . + \tan^{- 1} \left( \frac{1}{1 + n . \left( n + 1 \right)} \right) = \tan^{- 1} \theta\] , then find the value of θ.
Solve for x : {xcos(cot-1 x) + sin(cot-1 x)}2 = `51/50`
The value of sin `["cos"^-1 (7/25)]` is ____________.
