Advertisements
Advertisements
प्रश्न
Evaluate the following:
`tan^-1(tan4)`
उत्तर
We know that
`tan^-1(tantheta)=theta, -pi/2<theta<pi/2`
We have
`tan^-1(tan4)=tan^-1[tan(-pi+4)]`
= 4 - π
APPEARS IN
संबंधित प्रश्न
Solve the equation for x:sin−1x+sin−1(1−x)=cos−1x
Find the domain of definition of `f(x)=cos^-1(x^2-4)`
Find the principal values of the following:
`cos^-1(-sqrt3/2)`
Find the principal values of the following:
`cos^-1(tan (3pi)/4)`
`sin^-1(sin pi/6)`
`sin^-1(sin (13pi)/7)`
Evaluate the following:
`cos^-1{cos(-pi/4)}`
Evaluate the following:
`tan^-1(tan (7pi)/6)`
Evaluate the following:
`sec^-1(sec (5pi)/4)`
Evaluate the following:
`sec^-1(sec (9pi)/5)`
Evaluate the following:
`\text(cosec)^-1(\text{cosec} pi/4)`
Evaluate the following:
`cosec^-1{cosec (-(9pi)/4)}`
Write the following in the simplest form:
`tan^-1{(sqrt(1+x^2)-1)/x},x !=0`
Prove the following result
`tan(cos^-1 4/5+tan^-1 2/3)=17/6`
Evaluate:
`cosec{cot^-1(-12/5)}`
Evaluate:
`cot(tan^-1a+cot^-1a)`
`sin(sin^-1 1/5+cos^-1x)=1`
Solve the following equation for x:
cot−1x − cot−1(x + 2) =`pi/12`, x > 0
Solve `cos^-1sqrt3x+cos^-1x=pi/2`
Evaluate the following:
`tan{2tan^-1 1/5-pi/4}`
Evaluate the following:
`sin(2tan^-1 2/3)+cos(tan^-1sqrt3)`
Find the value of the following:
`tan^-1{2cos(2sin^-1 1/2)}`
Solve the following equation for x:
`3sin^-1 (2x)/(1+x^2)-4cos^-1 (1-x^2)/(1+x^2)+2tan^-1 (2x)/(1-x^2)=pi/3`
If `sin^-1x+sin^-1y+sin^-1z=(3pi)/2,` then write the value of x + y + z.
Write the value of cos−1 (cos 1540°).
Write the value of cos−1 (cos 6).
Write the value ofWrite the value of \[2 \sin^{- 1} \frac{1}{2} + \cos^{- 1} \left( - \frac{1}{2} \right)\]
Write the value of cos−1 \[\left( \cos\frac{5\pi}{4} \right)\]
Write the principal value of `sin^-1(-1/2)`
Write the value of \[\sin^{- 1} \left( \sin\frac{3\pi}{5} \right)\]
Write the value of `cot^-1(-x)` for all `x in R` in terms of `cot^-1(x)`
If θ = sin−1 {sin (−600°)}, then one of the possible values of θ is
If x = a (2θ – sin 2θ) and y = a (1 – cos 2θ), find \[\frac{dy}{dx}\] When \[\theta = \frac{\pi}{3}\] .
If y = sin (sin x), prove that \[\frac{d^2 y}{d x^2} + \tan x \frac{dy}{dx} + y \cos^2 x = 0 .\]
If 2 tan−1 (cos θ) = tan−1 (2 cosec θ), (θ ≠ 0), then find the value of θ.
Solve for x : {xcos(cot-1 x) + sin(cot-1 x)}2 = `51/50`
The value of tan `("cos"^-1 4/5 + "tan"^-1 2/3) =`