Advertisements
Advertisements
प्रश्न
Evaluate the following:
`sin(1/2cos^-1 4/5)`
उत्तर
`sin(1/2cos^-1 4/5)=sin{1/2xx2sin^-1+-sqrt((1-4/5)/2)}` `[becausecos^-1x=2sin^-1+-sqrt((1-x)/2)]`
`=sin(sin^-1+-1/sqrt10)`
`=+-1/sqrt10`
APPEARS IN
संबंधित प्रश्न
Solve the equation for x:sin−1x+sin−1(1−x)=cos−1x
Solve the following for x:
`sin^(-1)(1-x)-2sin^-1 x=pi/2`
Prove that
`tan^(-1) [(sqrt(1+x)-sqrt(1-x))/(sqrt(1+x)+sqrt(1-x))]=pi/4-1/2 cos^(-1)x,-1/sqrt2<=x<=1`
Find the principal values of the following:
`cos^-1(-1/sqrt2)`
`sin^-1(sin3)`
Evaluate the following:
`cos^-1{cos ((4pi)/3)}`
Evaluate the following:
`tan^-1(tan (6pi)/7)`
Evaluate the following:
`tan^-1(tan (7pi)/6)`
Evaluate the following:
`tan^-1(tan4)`
Evaluate the following:
`sec^-1(sec (7pi)/3)`
Evaluate the following:
`cosec^-1(cosec (13pi)/6)`
Write the following in the simplest form:
`tan^-1sqrt((a-x)/(a+x)),-a<x<a`
Evaluate the following:
`sin(sec^-1 17/8)`
Evaluate:
`cot(sin^-1 3/4+sec^-1 4/3)`
Find the value of `tan^-1 (x/y)-tan^-1((x-y)/(x+y))`
Solve the following equation for x:
`tan^-1 x/2+tan^-1 x/3=pi/4, 0<x<sqrt6`
Solve the following equation for x:
`tan^-1((x-2)/(x-4))+tan^-1((x+2)/(x+4))=pi/4`
`tan^-1 1/7+2tan^-1 1/3=pi/4`
If `sin^-1 (2a)/(1+a^2)-cos^-1 (1-b^2)/(1+b^2)=tan^-1 (2x)/(1-x^2)`, then prove that `x=(a-b)/(1+ab)`
Find the value of the following:
`tan^-1{2cos(2sin^-1 1/2)}`
Solve the following equation for x:
`tan^-1 1/4+2tan^-1 1/5+tan^-1 1/6+tan^-1 1/x=pi/4`
Solve the following equation for x:
`tan^-1((2x)/(1-x^2))+cot^-1((1-x^2)/(2x))=(2pi)/3,x>0`
If x > 1, then write the value of sin−1 `((2x)/(1+x^2))` in terms of tan−1 x.
Write the range of tan−1 x.
Write the value of cos2 \[\left( \frac{1}{2} \cos^{- 1} \frac{3}{5} \right)\]
Show that \[\sin^{- 1} (2x\sqrt{1 - x^2}) = 2 \sin^{- 1} x\]
Write the value of \[\tan^{- 1} \left\{ 2\sin\left( 2 \cos^{- 1} \frac{\sqrt{3}}{2} \right) \right\}\]
Write the value of \[\cos\left( \sin^{- 1} x + \cos^{- 1} x \right), \left| x \right| \leq 1\]
The value of tan \[\left\{ \cos^{- 1} \frac{1}{5\sqrt{2}} - \sin^{- 1} \frac{4}{\sqrt{17}} \right\}\] is
If sin−1 x − cos−1 x = `pi/6` , then x =
sin \[\left\{ 2 \cos^{- 1} \left( \frac{- 3}{5} \right) \right\}\] is equal to
The value of sin \[\left( \frac{1}{4} \sin^{- 1} \frac{\sqrt{63}}{8} \right)\] is
If tan−1 (cot θ) = 2 θ, then θ =
The domain of \[\cos^{- 1} \left( x^2 - 4 \right)\] is
If x = a (2θ – sin 2θ) and y = a (1 – cos 2θ), find \[\frac{dy}{dx}\] When \[\theta = \frac{\pi}{3}\] .
Prove that : \[\tan^{- 1} \left( \frac{\sqrt{1 + x^2} + \sqrt{1 - x^2}}{\sqrt{1 + x^2} - \sqrt{1 - x^2}} \right) = \frac{\pi}{4} + \frac{1}{2} \cos^{- 1} x^2 ; 1 < x < 1\].
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.
Find the domain of `sec^(-1)(3x-1)`.
The value of sin `["cos"^-1 (7/25)]` is ____________.