Advertisements
Advertisements
Question
Write the value of sin (cot−1 x).
Advertisements
Solution
We know
\[\cot^{- 1} x = \tan^{- 1} \frac{1}{x}\]
Now, we have
\[\sin\left( \cot^{- 1} x \right) = \sin\left( \tan^{- 1} \frac{1}{x} \right)\]
\[ = \sin\left[ \sin^{- 1} \left( \frac{\frac{1}{x}}{\sqrt{1 + \frac{1}{x^2}}} \right) \right] \left[ \because \tan^{- 1} x = \sin^{- 1} \left( \frac{x}{\sqrt{1 + x^2}} \right) \right]\]
\[ = \sin\left[ \sin^{- 1} \left( \frac{\frac{1}{x}}{\frac{\sqrt{x^2 + 1}}{x}} \right) \right]\]
\[ = \sin\left( \sin^{- 1} \frac{1}{\sqrt{x^2 + 1}} \right)\]
\[ = \frac{1}{\sqrt{x^2 + 1}} \left[ \because \sin\left( \sin^{- 1} x = x \right) \right]\]
Hence,
\[\sin\left( \cot^{- 1} x \right) = \frac{1}{\sqrt{x^2 - 1}}\]
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 (tan−1x)2 + (cot−1x)2 = 5π2/8, then find x.
If a line makes angles 90° and 60° respectively with the positive directions of x and y axes, find the angle which it makes with the positive direction of z-axis.
Find the domain of `f(x)=cos^-1x+cosx.`
`sin^-1(sin (5pi)/6)`
`sin^-1(sin12)`
`sin^-1(sin2)`
Evaluate the following:
`cos^-1{cos (5pi)/4}`
Evaluate the following:
`cos^-1(cos5)`
Evaluate the following:
`tan^-1(tan1)`
Evaluate the following:
`sec^-1(sec (5pi)/4)`
Evaluate the following:
`\text(cosec)^-1(\text{cosec} pi/4)`
Evaluate the following:
`cosec^-1(cosec (6pi)/5)`
Evaluate the following:
`sec(sin^-1 12/13)`
Evaluate:
`cosec{cot^-1(-12/5)}`
Evaluate:
`cos(sec^-1x+\text(cosec)^-1x)`,|x|≥1
If `sin^-1x+sin^-1y=pi/3` and `cos^-1x-cos^-1y=pi/6`, find the values of x and y.
`sin(sin^-1 1/5+cos^-1x)=1`
Find the value of `tan^-1 (x/y)-tan^-1((x-y)/(x+y))`
Solve the following equation for x:
tan−1(x + 2) + tan−1(x − 2) = tan−1 `(8/79)`, x > 0
`tan^-1 2/3=1/2tan^-1 12/5`
`2tan^-1 1/5+tan^-1 1/8=tan^-1 4/7`
`2tan^-1(1/2)+tan^-1(1/7)=tan^-1(31/17)`
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.`
Write the value of tan−1x + tan−1 `(1/x)`for x > 0.
Write the value of cos−1 (cos 1540°).
Write the value of \[\tan^{- 1} \frac{a}{b} - \tan^{- 1} \left( \frac{a - b}{a + b} \right)\]
Write the value of cos−1 \[\left( \cos\frac{5\pi}{4} \right)\]
Wnte the value of the expression \[\tan\left( \frac{\sin^{- 1} x + \cos^{- 1} x}{2} \right), \text { when } x = \frac{\sqrt{3}}{2}\]
2 tan−1 {cosec (tan−1 x) − tan (cot−1 x)} is equal to
The positive integral solution of the equation
\[\tan^{- 1} x + \cos^{- 1} \frac{y}{\sqrt{1 + y^2}} = \sin^{- 1} \frac{3}{\sqrt{10}}\text{ is }\]
If x < 0, y < 0 such that xy = 1, then tan−1 x + tan−1 y equals
If θ = sin−1 {sin (−600°)}, then one of the possible values of θ is
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 4 cos−1 x + sin−1 x = π, then the value of x is
Find : \[\int\frac{2 \cos x}{\left( 1 - \sin x \right) \left( 1 + \sin^2 x \right)}dx\] .
Prove that : \[\tan^{- 1} \left( \frac{\sqrt{1 + x^2} + \sqrt{1 - x^2}}{\sqrt{1 + x^2} - \sqrt{1 - x^2}} \right) = \frac{\pi}{4} + \frac{1}{2} \cos^{- 1} x^2 ; 1 < x < 1\].
If 2 tan−1 (cos θ) = tan−1 (2 cosec θ), (θ ≠ 0), then find the value of θ.
tanx is periodic with period ____________.
