Advertisements
Advertisements
Question
Evaluate the following:
`sec^-1(sec (2pi)/3)`
Advertisements
Solution
We know that
sec-1 (sec θ) = θ, [0, π/2) ∪ (π/2, π]
We have
`sec^-1(sec (2pi)/3)=(2pi)/3`
APPEARS IN
RELATED QUESTIONS
If `(sin^-1x)^2 + (sin^-1y)^2+(sin^-1z)^2=3/4pi^2,` find the value of x2 + y2 + z2
Find the principal values of the following:
`cos^-1(sin (4pi)/3)`
Find the principal values of the following:
`cos^-1(tan (3pi)/4)`
Evaluate the following:
`cos^-1{cos ((4pi)/3)}`
Evaluate the following:
`cos^-1(cos3)`
Evaluate the following:
`tan^-1(tan (7pi)/6)`
Evaluate the following:
`sec^-1(sec (9pi)/5)`
Evaluate the following:
`cosec^-1(cosec (3pi)/4)`
Write the following in the simplest form:
`tan^-1{(sqrt(1+x^2)-1)/x},x !=0`
Write the following in the simplest form:
`sin^-1{(sqrt(1+x)+sqrt(1-x))/2},0<x<1`
Evaluate the following:
`cot(cos^-1 3/5)`
Prove the following result
`cos(sin^-1 3/5+cot^-1 3/2)=6/(5sqrt13)`
Evaluate: `sin{cos^-1(-3/5)+cot^-1(-5/12)}`
Evaluate:
`sin(tan^-1x+tan^-1 1/x)` for x > 0
If `cos^-1x + cos^-1y =pi/4,` find the value of `sin^-1x+sin^-1y`
Find the value of `tan^-1 (x/y)-tan^-1((x-y)/(x+y))`
Solve the following equation for x:
cot−1x − cot−1(x + 2) =`pi/12`, x > 0
If `cos^-1 x/2+cos^-1 y/3=alpha,` then prove that `9x^2-12xy cosa+4y^2=36sin^2a.`
Evaluate the following:
`sin(2tan^-1 2/3)+cos(tan^-1sqrt3)`
`sin^-1 4/5+2tan^-1 1/3=pi/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`
Prove that `2tan^-1(sqrt((a-b)/(a+b))tan theta/2)=cos^-1((a costheta+b)/(a+b costheta))`
Write the value of `sin^-1((-sqrt3)/2)+cos^-1((-1)/2)`
If `sin^-1x+sin^-1y+sin^-1z=(3pi)/2,` then write the value of x + y + z.
If −1 < x < 0, then write the value of `sin^-1((2x)/(1+x^2))+cos^-1((1-x^2)/(1+x^2))`
Write the range of tan−1 x.
Write the value of cos−1 (cos 6).
Evaluate: \[\sin^{- 1} \left( \sin\frac{3\pi}{5} \right)\]
If \[\tan^{- 1} (\sqrt{3}) + \cot^{- 1} x = \frac{\pi}{2},\] find x.
If \[\sin^{- 1} \left( \frac{1}{3} \right) + \cos^{- 1} x = \frac{\pi}{2},\] then find x.
Write the value of \[\tan\left( 2 \tan^{- 1} \frac{1}{5} \right)\]
If sin−1 x − cos−1 x = `pi/6` , then x =
\[\text{ If }\cos^{- 1} \frac{x}{3} + \cos^{- 1} \frac{y}{2} = \frac{\theta}{2}, \text{ then }4 x^2 - 12xy \cos\frac{\theta}{2} + 9 y^2 =\]
sin \[\left\{ 2 \cos^{- 1} \left( \frac{- 3}{5} \right) \right\}\] is equal to
If \[\cos^{- 1} x > \sin^{- 1} x\], then
The domain of \[\cos^{- 1} \left( x^2 - 4 \right)\] is
Find : \[\int\frac{2 \cos x}{\left( 1 - \sin x \right) \left( 1 + \sin^2 x \right)}dx\] .
The value of tan `("cos"^-1 4/5 + "tan"^-1 2/3) =`
The equation sin-1 x – cos-1 x = cos-1 `(sqrt3/2)` has ____________.
