Advertisements
Advertisements
प्रश्न
Evaluate the following:
`tan^-1(tan1)`
Advertisements
उत्तर
We know that
`tan^-1(tantheta)=theta, -pi/2<theta<pi/2`
We have
`tan^-1(tan1)=1`
APPEARS IN
संबंधित प्रश्न
Solve for x:
`2tan^(-1)(cosx)=tan^(-1)(2"cosec" x)`
If tan-1x+tan-1y=π/4,xy<1, then write the value of x+y+xy.
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(-sqrt3/2)`
Find the principal values of the following:
`cos^-1(tan (3pi)/4)`
`sin^-1(sin (5pi)/6)`
Evaluate the following:
`cosec^-1(cosec (6pi)/5)`
Evaluate the following:
`cot^-1(cot (19pi)/6)`
Evaluate:
`cot{sec^-1(-13/5)}`
If `sin^-1x+sin^-1y=pi/3` and `cos^-1x-cos^-1y=pi/6`, find the values of x and y.
Evaluate the following:
`tan{2tan^-1 1/5-pi/4}`
`tan^-1 1/4+tan^-1 2/9=1/2cos^-1 3/2=1/2sin^-1(4/5)`
`sin^-1 4/5+2tan^-1 1/3=pi/2`
If `sin^-1 (2a)/(1+a^2)+sin^-1 (2b)/(1+b^2)=2tan^-1x,` Prove that `x=(a+b)/(1-ab).`
Show that `2tan^-1x+sin^-1 (2x)/(1+x^2)` is constant for x ≥ 1, find that constant.
Prove that:
`tan^-1 (2ab)/(a^2-b^2)+tan^-1 (2xy)/(x^2-y^2)=tan^-1 (2alphabeta)/(alpha^2-beta^2),` where `alpha=ax-by and beta=ay+bx.`
If x < 0, then write the value of cos−1 `((1-x^2)/(1+x^2))` in terms of tan−1 x.
If tan−1 x + tan−1 y = `pi/4`, then write the value of x + y + xy.
Write the value of cos−1 \[\left( \cos\frac{5\pi}{4} \right)\]
Show that \[\sin^{- 1} (2x\sqrt{1 - x^2}) = 2 \sin^{- 1} x\]
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 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 \[\sec^{- 1} \left( \frac{1}{2} \right)\]
Write the value of \[\cos^{- 1} \left( \cos\frac{14\pi}{3} \right)\]
Find the value of \[2 \sec^{- 1} 2 + \sin^{- 1} \left( \frac{1}{2} \right)\]
If \[\cos\left( \sin^{- 1} \frac{2}{5} + \cos^{- 1} x \right) = 0\], find the value of x.
Find the value of \[\tan^{- 1} \left( \tan\frac{9\pi}{8} \right)\]
The number of solutions of the equation \[\tan^{- 1} 2x + \tan^{- 1} 3x = \frac{\pi}{4}\] is
If α = \[\tan^{- 1} \left( \tan\frac{5\pi}{4} \right) \text{ and }\beta = \tan^{- 1} \left( - \tan\frac{2\pi}{3} \right)\] , then
\[\text{ If } u = \cot^{- 1} \sqrt{\tan \theta} - \tan^{- 1} \sqrt{\tan \theta}\text{ then }, \tan\left( \frac{\pi}{4} - \frac{u}{2} \right) =\]
If \[\cos^{- 1} \frac{x}{2} + \cos^{- 1} \frac{y}{3} = \theta,\] then 9x2 − 12xy cos θ + 4y2 is equal to
If \[3\sin^{- 1} \left( \frac{2x}{1 + x^2} \right) - 4 \cos^{- 1} \left( \frac{1 - x^2}{1 + x^2} \right) + 2 \tan^{- 1} \left( \frac{2x}{1 - x^2} \right) = \frac{\pi}{3}\] is equal to
The value of sin \[\left( \frac{1}{4} \sin^{- 1} \frac{\sqrt{63}}{8} \right)\] is
Find the real solutions of the equation
`tan^-1 sqrt(x(x + 1)) + sin^-1 sqrt(x^2 + x + 1) = pi/2`
The period of the function f(x) = tan3x is ____________.
