Advertisements
Advertisements
प्रश्न
Evaluate the following:
`sec^-1(sec (5pi)/4)`
उत्तर
We know that
sec-1 (sec θ) = θ, [0, π/2) ∪ (π/2, π]
We have
`sec^-1(sec (5pi)/4)=sec^-1[sec(2pi-(3pi)/4)]`
`=sec^-1[sec((3pi)/4)]`
`=(3pi)/4`
APPEARS IN
संबंधित प्रश्न
Solve the equation for x:sin−1x+sin−1(1−x)=cos−1x
Solve for x:
`2tan^(-1)(cosx)=tan^(-1)(2"cosec" x)`
Solve the following for x:
`sin^(-1)(1-x)-2sin^-1 x=pi/2`
Show that:
`2 sin^-1 (3/5)-tan^-1 (17/31)=pi/4`
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(sin (7pi)/6)`
`sin^-1(sin (17pi)/8)`
Evaluate the following:
`cos^-1(cos12)`
Evaluate the following:
`sec^-1(sec (9pi)/5)`
Evaluate the following:
`sec^-1(sec (13pi)/4)`
Evaluate the following:
`sec^-1(sec (25pi)/6)`
Evaluate the following:
`\text(cosec)^-1(\text{cosec} pi/4)`
Evaluate the following:
`cosec^-1(cosec (6pi)/5)`
Evaluate the following:
`cot^-1(cot (19pi)/6)`
Write the following in the simplest form:
`cot^-1 a/sqrt(x^2-a^2),| x | > a`
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{2tan^-1sqrt((1-x)/(1+x))}`
Evaluate the following:
`sin(tan^-1 24/7)`
Evaluate:
`tan{cos^-1(-7/25)}`
If `sin^-1x+sin^-1y=pi/3` and `cos^-1x-cos^-1y=pi/6`, find the values of x and y.
Solve the following equation for x:
tan−1`((1-x)/(1+x))-1/2` tan−1x = 0, where x > 0
Evaluate the following:
`tan 1/2(cos^-1 sqrt5/3)`
`tan^-1 1/7+2tan^-1 1/3=pi/4`
`2tan^-1 1/5+tan^-1 1/8=tan^-1 4/7`
Solve the following equation for x:
`cos^-1((x^2-1)/(x^2+1))+1/2tan^-1((2x)/(1-x^2))=(2x)/3`
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 sin−1 (sin 1550°).
Write the value of cos−1 (cos 350°) − sin−1 (sin 350°)
Write the value of cos−1 (cos 6).
Find the value of \[\cos^{- 1} \left( \cos\frac{13\pi}{6} \right)\]
The number of solutions of the equation \[\tan^{- 1} 2x + \tan^{- 1} 3x = \frac{\pi}{4}\] is
If θ = sin−1 {sin (−600°)}, then one of the possible values of θ is
If x > 1, then \[2 \tan^{- 1} x + \sin^{- 1} \left( \frac{2x}{1 + x^2} \right)\] is equal to
If y = sin (sin x), prove that \[\frac{d^2 y}{d x^2} + \tan x \frac{dy}{dx} + y \cos^2 x = 0 .\]
Prove that : \[\tan^{- 1} \left( \frac{\sqrt{1 + x^2} + \sqrt{1 - x^2}}{\sqrt{1 + x^2} - \sqrt{1 - x^2}} \right) = \frac{\pi}{4} + \frac{1}{2} \cos^{- 1} x^2 ; 1 < x < 1\].
Find the value of x, if tan `[sec^(-1) (1/x) ] = sin ( tan^(-1) 2) , x > 0 `.