Advertisements
Advertisements
प्रश्न
Evaluate the following:
`tan^-1(tan2)`
उत्तर
We know that
`tan^-1(tantheta)=theta, -pi/2<theta<pi/2`
We have
`tan^-1(tan2)=tan^-1[tan(-pi+2)]`
= 2 - π
APPEARS IN
संबंधित प्रश्न
Solve for x:
`2tan^(-1)(cosx)=tan^(-1)(2"cosec" x)`
If a line makes angles 90° and 60° respectively with the positive directions of x and y axes, find the angle which it makes with the positive direction of z-axis.
Find the principal values of the following:
`cos^-1(-1/sqrt2)`
`sin^-1(sin3)`
`sin^-1(sin4)`
Evaluate the following:
`cos^-1(cos4)`
Evaluate the following:
`tan^-1(tan pi/3)`
Evaluate the following:
`cosec^-1{cosec (-(9pi)/4)}`
Evaluate the following:
`cot^-1{cot (-(8pi)/3)}`
Write the following in the simplest form:
`cot^-1 a/sqrt(x^2-a^2),| x | > a`
Evaluate the following:
`sin(cos^-1 5/13)`
Evaluate the following:
`sin(tan^-1 24/7)`
Evaluate:
`sec{cot^-1(-5/12)}`
Evaluate:
`sin(tan^-1x+tan^-1 1/x)` for x > 0
Evaluate:
`cot(tan^-1a+cot^-1a)`
If `cot(cos^-1 3/5+sin^-1x)=0`, find the values of x.
`sin^-1x=pi/6+cos^-1x`
Prove the following result:
`sin^-1 12/13+cos^-1 4/5+tan^-1 63/16=pi`
Solve the following equation for x:
cot−1x − cot−1(x + 2) =`pi/12`, x > 0
Solve the following equation for x:
`tan^-1(2+x)+tan^-1(2-x)=tan^-1 2/3, where x< -sqrt3 or, x>sqrt3`
Solve the following:
`cos^-1x+sin^-1 x/2=π/6`
If `cos^-1 x/2+cos^-1 y/3=alpha,` then prove that `9x^2-12xy cosa+4y^2=36sin^2a.`
Solve `cos^-1sqrt3x+cos^-1x=pi/2`
Evaluate the following:
`tan{2tan^-1 1/5-pi/4}`
`2tan^-1(1/2)+tan^-1(1/7)=tan^-1(31/17)`
Solve the following equation for x:
`tan^-1 1/4+2tan^-1 1/5+tan^-1 1/6+tan^-1 1/x=pi/4`
Evaluate: \[\sin^{- 1} \left( \sin\frac{3\pi}{5} \right)\]
If \[\tan^{- 1} (\sqrt{3}) + \cot^{- 1} x = \frac{\pi}{2},\] find x.
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 principal value of `tan^-1sqrt3+cot^-1sqrt3`
Wnte the value of the expression \[\tan\left( \frac{\sin^{- 1} x + \cos^{- 1} x}{2} \right), \text { when } x = \frac{\sqrt{3}}{2}\]
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 α = \[\tan^{- 1} \left( \frac{\sqrt{3}x}{2y - x} \right), \beta = \tan^{- 1} \left( \frac{2x - y}{\sqrt{3}y} \right),\]
then α − β =
sin \[\left\{ 2 \cos^{- 1} \left( \frac{- 3}{5} \right) \right\}\] is equal to
The value of \[\sin\left( 2\left( \tan^{- 1} 0 . 75 \right) \right)\] is equal to
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 simplified form of `cos^-1 (3/5 cosx + 4/5 sin x)`, where x ∈ `[(-3pi)/4, pi/4]`
The value of sin `["cos"^-1 (7/25)]` is ____________.