Advertisements
Advertisements
Question
Evaluate the following:
`tan^-1(tan12)`
Advertisements
Solution
We know that
`tan^-1(tantheta)=theta, -pi/2<theta<pi/2`
We have
`tan^-1(tan12)=tan^-1[tan(-4pi+12)]`
= 12 - 4π
APPEARS IN
RELATED QUESTIONS
`sin^-1(sin (5pi)/6)`
Evaluate the following:
`cos^-1(cos5)`
Evaluate the following:
`tan^-1(tan (7pi)/6)`
Evaluate the following:
`sec^-1(sec pi/3)`
Evaluate the following:
`\text(cosec)^-1(\text{cosec} pi/4)`
Evaluate the following:
`cot^-1{cot ((21pi)/4)}`
Write the following in the simplest form:
`tan^-1sqrt((a-x)/(a+x)),-a<x<a`
Write the following in the simplest form:
`sin^-1{(sqrt(1+x)+sqrt(1-x))/2},0<x<1`
Evaluate the following:
`sin(sin^-1 7/25)`
Evaluate the following:
`tan(cos^-1 8/17)`
Evaluate:
`sec{cot^-1(-5/12)}`
Evaluate:
`cot{sec^-1(-13/5)}`
`sin(sin^-1 1/5+cos^-1x)=1`
Solve the following equation for x:
`tan^-1 2x+tan^-1 3x = npi+(3pi)/4`
Solve the following equation for x:
cot−1x − cot−1(x + 2) =`pi/12`, x > 0
`sin^-1 5/13+cos^-1 3/5=tan^-1 63/16`
Solve the following:
`sin^-1x+sin^-1 2x=pi/3`
Evaluate the following:
`tan{2tan^-1 1/5-pi/4}`
Prove that:
`2sin^-1 3/5=tan^-1 24/7`
`tan^-1 2/3=1/2tan^-1 12/5`
`2tan^-1 3/4-tan^-1 17/31=pi/4`
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)`
Prove that `2tan^-1(sqrt((a-b)/(a+b))tan theta/2)=cos^-1((a costheta+b)/(a+b costheta))`
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`
If `sin^-1x+sin^-1y+sin^-1z=(3pi)/2,` then write the value of x + y + z.
Write the value of sin−1
\[\left( \sin( -{600}°) \right)\].
Write the value of cos−1 (cos 350°) − sin−1 (sin 350°)
Evaluate: \[\sin^{- 1} \left( \sin\frac{3\pi}{5} \right)\]
Write the principal value of \[\cos^{- 1} \left( \cos\frac{2\pi}{3} \right) + \sin^{- 1} \left( \sin\frac{2\pi}{3} \right)\]
Write the value of \[\tan^{- 1} \left\{ 2\sin\left( 2 \cos^{- 1} \frac{\sqrt{3}}{2} \right) \right\}\]
Write the principal value of \[\cos^{- 1} \left( \cos680^\circ \right)\]
If sin−1 x − cos−1 x = `pi/6` , then x =
Let f (x) = `e^(cos^-1){sin(x+pi/3}.`
Then, f (8π/9) =
sin \[\left\{ 2 \cos^{- 1} \left( \frac{- 3}{5} \right) \right\}\] is equal to
The value of sin \[\left( \frac{1}{4} \sin^{- 1} \frac{\sqrt{63}}{8} \right)\] is
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 `.
The value of tan `("cos"^-1 4/5 + "tan"^-1 2/3) =`
