Advertisements
Advertisements
प्रश्न
Solve `cos^-1sqrt3x+cos^-1x=pi/2`
Advertisements
उत्तर
`cos^-1sqrt3x+cos^-1x=pi/2`
⇒`cos^-1[sqrt3x xx x-sqrt(1-(sqrt3x)^2)sqrt(1-x^2)]=pi/2` `[becausecos^-1x+cos^-1y=cos^-1(xy-sqrt(1-x^2)sqrt(1-y^2)]`
⇒ `cos^-1[sqrt3x^2-sqrt(1-3x^2)sqrt(1-x^2)]=pi/2`
⇒ `sqrt3x^2=sqrt(1-3x^2)sqrt(1-x^2)=cos pi/2`
⇒ `sqrt3x^2=sqrt(1-3x^2)sqrt(1-x^2)`
⇒ `3x^4=(1-3x^2)(1-x^2)`
⇒ `3x^4=1-3x^2+3x^4-x^2`
⇒ `4x^2=1`
⇒ `x^2=1/4`
⇒ `x=+-1/2`
APPEARS IN
संबंधित प्रश्न
Prove that :
`2 tan^-1 (sqrt((a-b)/(a+b))tan(x/2))=cos^-1 ((a cos x+b)/(a+b cosx))`
If sin [cot−1 (x+1)] = cos(tan−1x), then find x.
If tan-1x+tan-1y=π/4,xy<1, then write the value of x+y+xy.
Find the principal values of the following:
`cos^-1(sin (4pi)/3)`
Evaluate the following:
`tan^-1(tan pi/3)`
Evaluate the following:
`tan^-1(tan2)`
Evaluate the following:
`sec^-1(sec pi/3)`
Evaluate the following:
`sec^-1(sec (5pi)/4)`
Evaluate the following:
`sec^-1(sec (13pi)/4)`
Evaluate the following:
`cosec^-1(cosec (6pi)/5)`
Write the following in the simplest form:
`tan^-1{x+sqrt(1+x^2)},x in R `
Write the following in the simplest form:
`tan^-1{sqrt(1+x^2)-x},x in R`
Write the following in the simplest form:
`tan^-1sqrt((a-x)/(a+x)),-a<x<a`
Write the following in the simplest form:
`tan^-1(x/(a+sqrt(a^2-x^2))),-a<x<a`
Evaluate:
`sin(tan^-1x+tan^-1 1/x)` for x < 0
Solve the following equation for x:
`tan^-1((x-2)/(x-4))+tan^-1((x+2)/(x+4))=pi/4`
Sum the following series:
`tan^-1 1/3+tan^-1 2/9+tan^-1 4/33+...+tan^-1 (2^(n-1))/(1+2^(2n-1))`
Evaluate: `cos(sin^-1 3/5+sin^-1 5/13)`
Solve the following:
`sin^-1x+sin^-1 2x=pi/3`
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)`
Solve the following equation for x:
`tan^-1((2x)/(1-x^2))+cot^-1((1-x^2)/(2x))=(2pi)/3,x>0`
If x < 0, then write the value of cos−1 `((1-x^2)/(1+x^2))` in terms of tan−1 x.
Write the value of cos−1 (cos 350°) − sin−1 (sin 350°)
Write the value of cos−1 (cos 6).
Write the value of sin \[\left\{ \frac{\pi}{3} - \sin^{- 1} \left( - \frac{1}{2} \right) \right\}\]
Wnte the value of the expression \[\tan\left( \frac{\sin^{- 1} x + \cos^{- 1} x}{2} \right), \text { when } x = \frac{\sqrt{3}}{2}\]
Wnte the value of\[\cos\left( \frac{\tan^{- 1} x + \cot^{- 1} x}{3} \right), \text{ when } x = - \frac{1}{\sqrt{3}}\]
If \[\cos\left( \sin^{- 1} \frac{2}{5} + \cos^{- 1} x \right) = 0\], find the value of x.
If sin−1 x − cos−1 x = `pi/6` , then x =
sin\[\left[ \cot^{- 1} \left\{ \tan\left( \cos^{- 1} x \right) \right\} \right]\] is equal to
If α = \[\tan^{- 1} \left( \tan\frac{5\pi}{4} \right) \text{ and }\beta = \tan^{- 1} \left( - \tan\frac{2\pi}{3} \right)\] , then
If tan−1 3 + tan−1 x = tan−1 8, then x =
If θ = sin−1 {sin (−600°)}, then one of the possible values of θ is
If \[3\sin^{- 1} \left( \frac{2x}{1 + x^2} \right) - 4 \cos^{- 1} \left( \frac{1 - x^2}{1 + x^2} \right) + 2 \tan^{- 1} \left( \frac{2x}{1 - x^2} \right) = \frac{\pi}{3}\] is equal to
Find : \[\int\frac{2 \cos x}{\left( 1 - \sin x \right) \left( 1 + \sin^2 x \right)}dx\] .
Prove that : \[\tan^{- 1} \left( \frac{\sqrt{1 + x^2} + \sqrt{1 - x^2}}{\sqrt{1 + x^2} - \sqrt{1 - x^2}} \right) = \frac{\pi}{4} + \frac{1}{2} \cos^{- 1} x^2 ; 1 < x < 1\].
Solve for x : {xcos(cot-1 x) + sin(cot-1 x)}2 = `51/50`
