Advertisements
Advertisements
प्रश्न
Show that `2tan^-1x+sin^-1 (2x)/(1+x^2)` is constant for x ≥ 1, find that constant.
उत्तर
We have
`2tan^-1x+sin^-1 ((2x)/(1+x^2))`
(1) For x > 1,
`=2tan^-1x+sin^-1 ((2x)/(1+x^2))`
`=pi-sin^-1((2x)/(1+x^2))+sin^-1((2x)/(1+x^2))` `[because 2tan^-1x=pi - sin^-1((2x)/(1+x^2)),x>1]`
`=pi`
(2) For x = 1,
`=2tan^-1x+sin^-1 ((2x)/(1+x^2))`
`=2tan^-1(1)+sin^-1((2(1))/(1+(1)^2))`
`=2tan^-1(1)+sin^-1(1)`
`=2(pi/4)+pi/2`
= π
APPEARS IN
संबंधित प्रश्न
Prove that
`tan^(-1) [(sqrt(1+x)-sqrt(1-x))/(sqrt(1+x)+sqrt(1-x))]=pi/4-1/2 cos^(-1)x,-1/sqrt2<=x<=1`
`sin^-1(sin pi/6)`
`sin^-1(sin (13pi)/7)`
Evaluate the following:
`cos^-1(cos3)`
Evaluate the following:
`tan^-1(tan (6pi)/7)`
Evaluate the following:
`cosec^-1(cosec (11pi)/6)`
Evaluate the following:
`cosec^-1(cosec (13pi)/6)`
Evaluate the following:
`cot^-1(cot (4pi)/3)`
Evaluate the following:
`cot^-1(cot (9pi)/4)`
Write the following in the simplest form:
`cot^-1 a/sqrt(x^2-a^2),| x | > a`
Write the following in the simplest form:
`tan^-1{x+sqrt(1+x^2)},x in R `
Write the following in the simplest form:
`tan^-1{(sqrt(1+x^2)-1)/x},x !=0`
Evaluate the following:
`cos(tan^-1 24/7)`
Prove the following result
`tan(cos^-1 4/5+tan^-1 2/3)=17/6`
Evaluate:
`tan{cos^-1(-7/25)}`
If `cos^-1x + cos^-1y =pi/4,` find the value of `sin^-1x+sin^-1y`
Solve the following equation for x:
`tan^-1 2x+tan^-1 3x = npi+(3pi)/4`
`sin^-1 63/65=sin^-1 5/13+cos^-1 3/5`
If `cos^-1 x/2+cos^-1 y/3=alpha,` then prove that `9x^2-12xy cosa+4y^2=36sin^2a.`
Prove that: `cos^-1 4/5+cos^-1 12/13=cos^-1 33/65`
`tan^-1 2/3=1/2tan^-1 12/5`
`2tan^-1 1/5+tan^-1 1/8=tan^-1 4/7`
`4tan^-1 1/5-tan^-1 1/239=pi/4`
Solve the following equation for x:
`tan^-1((2x)/(1-x^2))+cot^-1((1-x^2)/(2x))=(2pi)/3,x>0`
If x > 1, then write the value of sin−1 `((2x)/(1+x^2))` in terms of tan−1 x.
Write the value of
\[\cos^{- 1} \left( \frac{1}{2} \right) + 2 \sin^{- 1} \left( \frac{1}{2} \right)\].
Write the value of cos−1 (cos 350°) − sin−1 (sin 350°)
If tan−1 x + tan−1 y = `pi/4`, then write the value of x + y + xy.
Write the value of \[\tan^{- 1} \frac{a}{b} - \tan^{- 1} \left( \frac{a - b}{a + b} \right)\]
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 `tan^-1sqrt3+cot^-1sqrt3`
If sin−1 x − cos−1 x = `pi/6` , then x =
\[\text{ If } u = \cot^{- 1} \sqrt{\tan \theta} - \tan^{- 1} \sqrt{\tan \theta}\text{ then }, \tan\left( \frac{\pi}{4} - \frac{u}{2} \right) =\]
\[\tan^{- 1} \frac{1}{11} + \tan^{- 1} \frac{2}{11}\] is equal to
It \[\tan^{- 1} \frac{x + 1}{x - 1} + \tan^{- 1} \frac{x - 1}{x} = \tan^{- 1}\] (−7), then the value of x is
If tan−1 (cot θ) = 2 θ, then θ =
Find the value of x, if tan `[sec^(-1) (1/x) ] = sin ( tan^(-1) 2) , x > 0 `.