Advertisements
Advertisements
Question
Write the value of \[\tan^{- 1} \left( \frac{1}{x} \right)\] for x < 0 in terms of `cot^-1x`
Advertisements
Solution
\[\tan^{- 1} \left( \frac{1}{x} \right) = \tan^{- 1} \left( - \frac{1}{x} \right)\text{ for } x < 0\]
\[ = - \tan^{- 1} \left( \frac{1}{x} \right)\]
\[ = - \cot^{- 1} x\]
\[ = - \left( \pi - \cot^{- 1} x \right)\]
\[ = - \pi + \cot^{- 1} x\]
APPEARS IN
RELATED QUESTIONS
Find the value of the following: `tan(1/2)[sin^(-1)((2x)/(1+x^2))+cos^(-1)((1-y^2)/(1+y^2))],|x| <1,y>0 and xy <1`
If `cos^-1( x/a) +cos^-1 (y/b)=alpha` , prove that `x^2/a^2-2(xy)/(ab) cos alpha +y^2/b^2=sin^2alpha`
Show that:
`2 sin^-1 (3/5)-tan^-1 (17/31)=pi/4`
If `tan^(-1)((x-2)/(x-4)) +tan^(-1)((x+2)/(x+4))=pi/4` ,find the value of x
Find the domain of `f(x) =2cos^-1 2x+sin^-1x.`
`sin^-1(sin (5pi)/6)`
`sin^-1(sin4)`
Evaluate the following:
`tan^-1(tan (7pi)/6)`
Evaluate the following:
`tan^-1(tan2)`
Evaluate the following:
`cosec^-1(cosec (6pi)/5)`
Write the following in the simplest form:
`tan^-1{x+sqrt(1+x^2)},x in R `
Write the following in the simplest form:
`sin^-1{(x+sqrt(1-x^2))/sqrt2},-1<x<1`
Evaluate the following:
`cosec(cos^-1 3/5)`
Evaluate the following:
`sec(sin^-1 12/13)`
Evaluate the following:
`tan(cos^-1 8/17)`
Prove the following result
`sin(cos^-1 3/5+sin^-1 5/13)=63/65`
Evaluate: `sin{cos^-1(-3/5)+cot^-1(-5/12)}`
Evaluate:
`sin(tan^-1x+tan^-1 1/x)` for x < 0
`sin^-1x=pi/6+cos^-1x`
`tan^-1x+2cot^-1x=(2x)/3`
Solve the following equation for x:
`tan^-1((1-x)/(1+x))-1/2 tan^-1x` = 0, where 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`
Evaluate: `cos(sin^-1 3/5+sin^-1 5/13)`
`(9pi)/8-9/4sin^-1 1/3=9/4sin^-1 (2sqrt2)/3`
Solve the following:
`cos^-1x+sin^-1 x/2=π/6`
Solve the equation `cos^-1 a/x-cos^-1 b/x=cos^-1 1/b-cos^-1 1/a`
Prove that: `cos^-1 4/5+cos^-1 12/13=cos^-1 33/65`
`2tan^-1(1/2)+tan^-1(1/7)=tan^-1(31/17)`
Prove that
`sin{tan^-1 (1-x^2)/(2x)+cos^-1 (1-x^2)/(2x)}=1`
If `sin^-1 (2a)/(1+a^2)+sin^-1 (2b)/(1+b^2)=2tan^-1x,` Prove that `x=(a+b)/(1-ab).`
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.
Write the value of cos\[\left( 2 \sin^{- 1} \frac{1}{3} \right)\]
Write the principal value of `tan^-1sqrt3+cot^-1sqrt3`
The number of real solutions of the equation \[\sqrt{1 + \cos 2x} = \sqrt{2} \sin^{- 1} (\sin x), - \pi \leq x \leq \pi\]
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 =
The value of sin \[\left( \frac{1}{4} \sin^{- 1} \frac{\sqrt{63}}{8} \right)\] is
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
Find the value of x, if tan `[sec^(-1) (1/x) ] = sin ( tan^(-1) 2) , x > 0 `.
The value of sin `["cos"^-1 (7/25)]` is ____________.
