Advertisements
Advertisements
प्रश्न
Evaluate the following:
`cos^-1{cos ((4pi)/3)}`
उत्तर
We know
`cos^-1(costheta)=thetaif 0<=theta<=pi`
We have
`cos^-1{cos ((4pi)/3)}=cos^-1{cos(2pi-(2pi)/3)}`
`=cos^-1{cos((2pi)/3)}`
`=(2pi)/3`
APPEARS IN
संबंधित प्रश्न
Write the value of `tan(2tan^(-1)(1/5))`
Solve the equation for x:sin−1x+sin−1(1−x)=cos−1x
Prove that :
`2 tan^-1 (sqrt((a-b)/(a+b))tan(x/2))=cos^-1 ((a cos x+b)/(a+b cosx))`
Solve for x:
`2tan^(-1)(cosx)=tan^(-1)(2"cosec" x)`
`sin^-1{(sin - (17pi)/8)}`
Evaluate the following:
`cos^-1{cos(-pi/4)}`
Evaluate the following:
`tan^-1(tan4)`
Evaluate the following:
`sec^-1(sec (2pi)/3)`
Evaluate the following:
`sin(sec^-1 17/8)`
Evaluate the following:
`sec(sin^-1 12/13)`
Solve: `cos(sin^-1x)=1/6`
Evaluate:
`cos(tan^-1 3/4)`
If `cos^-1x + cos^-1y =pi/4,` find the value of `sin^-1x+sin^-1y`
`sin^-1x=pi/6+cos^-1x`
`4sin^-1x=pi-cos^-1x`
Solve the following equation for x:
`tan^-1 2x+tan^-1 3x = npi+(3pi)/4`
Solve the following equation for x:
tan−1(x + 1) + tan−1(x − 1) = tan−1`8/31`
Solve the following equation for x:
tan−1(x + 2) + tan−1(x − 2) = tan−1 `(8/79)`, x > 0
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-1)+tan^-1 (x+2)/(x+1)=pi/4`
Evaluate: `cos(sin^-1 3/5+sin^-1 5/13)`
`sin^-1 63/65=sin^-1 5/13+cos^-1 3/5`
Evaluate the following:
`tan{2tan^-1 1/5-pi/4}`
Evaluate the following:
`sin(2tan^-1 2/3)+cos(tan^-1sqrt3)`
`tan^-1 1/4+tan^-1 2/9=1/2cos^-1 3/2=1/2sin^-1(4/5)`
Find the value of the following:
`tan^-1{2cos(2sin^-1 1/2)}`
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.
What is the value of cos−1 `(cos (2x)/3)+sin^-1(sin (2x)/3)?`
Write the value of sin (cot−1 x).
Write the value of cos−1 (cos 1540°).
If 4 sin−1 x + cos−1 x = π, then what is the value of x?
If \[\cos^{- 1} \frac{x}{a} + \cos^{- 1} \frac{y}{b} = \alpha, then\frac{x^2}{a^2} - \frac{2xy}{ab}\cos \alpha + \frac{y^2}{b^2} = \]
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 \[\cos^{- 1} \frac{x}{2} + \cos^{- 1} \frac{y}{3} = \theta,\] then 9x2 − 12xy cos θ + 4y2 is equal to
If \[\sin^{- 1} \left( \frac{2a}{1 - a^2} \right) + \cos^{- 1} \left( \frac{1 - a^2}{1 + a^2} \right) = \tan^{- 1} \left( \frac{2x}{1 - x^2} \right),\text{ where }a, x \in \left( 0, 1 \right)\] , then, the value of x is
If x > 1, then \[2 \tan^{- 1} x + \sin^{- 1} \left( \frac{2x}{1 + x^2} \right)\] is equal to
The value of sin `["cos"^-1 (7/25)]` is ____________.
