Advertisements
Advertisements
प्रश्न
Evaluate the following:
`tan^-1(tan12)`
Advertisements
उत्तर
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
संबंधित प्रश्न
Solve the following for x :
`tan^(-1)((x-2)/(x-3))+tan^(-1)((x+2)/(x+3))=pi/4,|x|<1`
`sin^-1(sin3)`
Evaluate the following:
`cos^-1{cos(-pi/4)}`
Evaluate the following:
`cos^-1(cos12)`
Evaluate the following:
`tan^-1(tan4)`
Evaluate the following:
`cosec^-1(cosec (11pi)/6)`
Evaluate the following:
`sin(sin^-1 7/25)`
Evaluate the following:
`sec(sin^-1 12/13)`
Evaluate:
`cot(tan^-1a+cot^-1a)`
If `cot(cos^-1 3/5+sin^-1x)=0`, find the values of x.
`tan^-1x+2cot^-1x=(2x)/3`
Prove the following result:
`sin^-1 12/13+cos^-1 4/5+tan^-1 63/16=pi`
Evaluate the following:
`sin(1/2cos^-1 4/5)`
Evaluate the following:
`sin(2tan^-1 2/3)+cos(tan^-1sqrt3)`
`2tan^-1(1/2)+tan^-1(1/7)=tan^-1(31/17)`
Solve the following equation for x:
`tan^-1((x-2)/(x-1))+tan^-1((x+2)/(x+1))=pi/4`
Prove that `2tan^-1(sqrt((a-b)/(a+b))tan theta/2)=cos^-1((a costheta+b)/(a+b costheta))`
Write the difference between maximum and minimum values of sin−1 x for x ∈ [− 1, 1].
Evaluate sin
\[\left( \frac{1}{2} \cos^{- 1} \frac{4}{5} \right)\]
Write the value of cos−1 (cos 350°) − sin−1 (sin 350°)
Write the value of sin \[\left\{ \frac{\pi}{3} - \sin^{- 1} \left( - \frac{1}{2} \right) \right\}\]
Write the value of cos−1 \[\left( \cos\frac{5\pi}{4} \right)\]
Evaluate: \[\sin^{- 1} \left( \sin\frac{3\pi}{5} \right)\]
If \[\sin^{- 1} \left( \frac{1}{3} \right) + \cos^{- 1} x = \frac{\pi}{2},\] then find x.
Write the value of \[\tan\left( 2 \tan^{- 1} \frac{1}{5} \right)\]
Write the principal value of `tan^-1sqrt3+cot^-1sqrt3`
Write the principal value of \[\cos^{- 1} \left( \cos680^\circ \right)\]
If \[\tan^{- 1} \left( \frac{\sqrt{1 + x^2} - \sqrt{1 - x^2}}{\sqrt{1 + x^2} + \sqrt{1 - x^2}} \right)\] = α, then x2 =
If x < 0, y < 0 such that xy = 1, then tan−1 x + tan−1 y equals
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 =
\[\tan^{- 1} \frac{1}{11} + \tan^{- 1} \frac{2}{11}\] 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
If x > 1, then \[2 \tan^{- 1} x + \sin^{- 1} \left( \frac{2x}{1 + x^2} \right)\] is equal to
Find : \[\int\frac{2 \cos x}{\left( 1 - \sin x \right) \left( 1 + \sin^2 x \right)}dx\] .
Prove that : \[\cot^{- 1} \frac{\sqrt{1 + \sin x} + \sqrt{1 - \sin x}}{\sqrt{1 + \sin x} - \sqrt{1 - \sin x}} = \frac{x}{2}, 0 < x < \frac{\pi}{2}\] .
The value of tan `("cos"^-1 4/5 + "tan"^-1 2/3) =`
