Advertisements
Advertisements
प्रश्न
Evaluate:
`sin(tan^-1x+tan^-1 1/x)` for x > 0
उत्तर
`sin(tan^-1x+tan^-1 1/x)`
`=sin(tan^-1x+cot^-1x)` `[thereforetan^-1x=cot^-1 1/x]`
`=sin(pi/2)` `[thereforetan^-1x=cot^-1x=pi/2]`
= 1
APPEARS IN
संबंधित प्रश्न
Write the value of `tan(2tan^(-1)(1/5))`
Solve for x:
`2tan^(-1)(cosx)=tan^(-1)(2"cosec" x)`
Find the domain of `f(x) =2cos^-1 2x+sin^-1x.`
`sin^-1(sin (13pi)/7)`
`sin^-1(sin4)`
Evaluate the following:
`cos^-1(cos4)`
Evaluate the following:
`cosec^-1{cosec (-(9pi)/4)}`
Evaluate the following:
`cot^-1{cot ((21pi)/4)}`
Evaluate the following:
`sin(cos^-1 5/13)`
Evaluate the following:
`sin(tan^-1 24/7)`
Evaluate the following:
`sec(sin^-1 12/13)`
Evaluate the following:
`tan(cos^-1 8/17)`
Evaluate the following:
`cot(cos^-1 3/5)`
Prove the following result-
`tan^-1 63/16 = sin^-1 5/13 + cos^-1 3/5`
Solve the following equation for x:
`tan^-1((x-2)/(x-4))+tan^-1((x+2)/(x+4))=pi/4`
`2tan^-1(1/2)+tan^-1(1/7)=tan^-1(31/17)`
`4tan^-1 1/5-tan^-1 1/239=pi/4`
Find the value of the following:
`tan^-1{2cos(2sin^-1 1/2)}`
Solve the following equation for x:
`tan^-1((2x)/(1-x^2))+cot^-1((1-x^2)/(2x))=(2pi)/3,x>0`
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.`
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`
Write the value of tan−1x + tan−1 `(1/x)`for x > 0.
Write the value of sin (cot−1 x).
Write the value of cos\[\left( 2 \sin^{- 1} \frac{1}{3} \right)\]
Write the value of sin−1 \[\left( \cos\frac{\pi}{9} \right)\]
Write the value of tan−1\[\left\{ \tan\left( \frac{15\pi}{4} \right) \right\}\]
If \[\sin^{- 1} \left( \frac{1}{3} \right) + \cos^{- 1} x = \frac{\pi}{2},\] then find x.
Write the principal value of \[\sin^{- 1} \left\{ \cos\left( \sin^{- 1} \frac{1}{2} \right) \right\}\]
Write the value of `cot^-1(-x)` for all `x in R` in terms of `cot^-1(x)`
The positive integral solution of the equation
\[\tan^{- 1} x + \cos^{- 1} \frac{y}{\sqrt{1 + y^2}} = \sin^{- 1} \frac{3}{\sqrt{10}}\text{ is }\]
\[\text{ If } u = \cot^{- 1} \sqrt{\tan \theta} - \tan^{- 1} \sqrt{\tan \theta}\text{ then }, \tan\left( \frac{\pi}{4} - \frac{u}{2} \right) =\]
\[\text{ If }\cos^{- 1} \frac{x}{3} + \cos^{- 1} \frac{y}{2} = \frac{\theta}{2}, \text{ then }4 x^2 - 12xy \cos\frac{\theta}{2} + 9 y^2 =\]
Let f (x) = `e^(cos^-1){sin(x+pi/3}.`
Then, f (8π/9) =
If tan−1 3 + tan−1 x = tan−1 8, then x =
If \[\cos^{- 1} x > \sin^{- 1} x\], then
If tan−1 (cot θ) = 2 θ, then θ =
If \[\sin^{- 1} \left( \frac{2a}{1 - a^2} \right) + \cos^{- 1} \left( \frac{1 - a^2}{1 + a^2} \right) = \tan^{- 1} \left( \frac{2x}{1 - x^2} \right),\text{ where }a, x \in \left( 0, 1 \right)\] , then, the value of x is
The value of \[\sin\left( 2\left( \tan^{- 1} 0 . 75 \right) \right)\] is equal to
If x = a (2θ – sin 2θ) and y = a (1 – cos 2θ), find \[\frac{dy}{dx}\] When \[\theta = \frac{\pi}{3}\] .