Advertisements
Advertisements
प्रश्न
Write the following in the simplest form:
`sin{2tan^-1sqrt((1-x)/(1+x))}`
Advertisements
उत्तर
Let x = cos θ
Now,
`sin{2tan^-1sqrt((1-x)/(1+x))}=sin{2tan^-1sqrt((1-costheta)/(1+costheta))}`
`=sin{2tan^-1sqrt((2sin^2 theta/2)/(2cos^2 theta/2))}`
`=sin{2tan^-1(tan theta/2)}`
= sin θ
= sin (cos-1 x)
`=sin(sin^-1(sqrt(1-x^2)))`
`=sqrt(1-x^2)`
APPEARS IN
संबंधित प्रश्न
If sin [cot−1 (x+1)] = cos(tan−1x), then find x.
If tan-1x+tan-1y=π/4,xy<1, then write the value of x+y+xy.
`sin^-1(sin3)`
`sin^-1(sin2)`
Evaluate the following:
`cos^-1{cos ((4pi)/3)}`
Evaluate the following:
`tan^-1(tan (6pi)/7)`
Evaluate the following:
`sec^-1(sec (9pi)/5)`
Evaluate the following:
`sec^-1(sec (13pi)/4)`
Evaluate the following:
`cosec^-1(cosec (11pi)/6)`
Evaluate the following:
`cot^-1(cot (19pi)/6)`
Evaluate the following:
`sin(tan^-1 24/7)`
Evaluate the following:
`cosec(cos^-1 3/5)`
Prove the following result
`cos(sin^-1 3/5+cot^-1 3/2)=6/(5sqrt13)`
Evaluate:
`tan{cos^-1(-7/25)}`
Evaluate:
`cosec{cot^-1(-12/5)}`
Evaluate:
`cos(tan^-1 3/4)`
If `sin^-1x+sin^-1y=pi/3` and `cos^-1x-cos^-1y=pi/6`, find the values of x and y.
Prove the following result:
`tan^-1 1/4+tan^-1 2/9=sin^-1 1/sqrt5`
Solve the following equation for x:
`tan^-1 x/2+tan^-1 x/3=pi/4, 0<x<sqrt6`
Prove that:
`2sin^-1 3/5=tan^-1 24/7`
`tan^-1 2/3=1/2tan^-1 12/5`
Prove that
`tan^-1((1-x^2)/(2x))+cot^-1((1-x^2)/(2x))=pi/2`
If `sin^-1 (2a)/(1+a^2)+sin^-1 (2b)/(1+b^2)=2tan^-1x,` Prove that `x=(a+b)/(1-ab).`
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 tan−1x + tan−1 `(1/x)`for x > 0.
Write the value of tan−1\[\left\{ \tan\left( \frac{15\pi}{4} \right) \right\}\]
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.
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 =
The value of tan \[\left\{ \cos^{- 1} \frac{1}{5\sqrt{2}} - \sin^{- 1} \frac{4}{\sqrt{17}} \right\}\] is
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 \[\cos^{- 1} \frac{x}{2} + \cos^{- 1} \frac{y}{3} = \theta,\] then 9x2 − 12xy cos θ + 4y2 is equal to
The value of \[\sin\left( 2\left( \tan^{- 1} 0 . 75 \right) \right)\] is equal to
If \[\tan^{- 1} \left( \frac{1}{1 + 1 . 2} \right) + \tan^{- 1} \left( \frac{1}{1 + 2 . 3} \right) + . . . + \tan^{- 1} \left( \frac{1}{1 + n . \left( n + 1 \right)} \right) = \tan^{- 1} \theta\] , then find the value of θ.
Write the value of \[\cos^{- 1} \left( - \frac{1}{2} \right) + 2 \sin^{- 1} \left( \frac{1}{2} \right)\] .
Find the domain of `sec^(-1)(3x-1)`.
tanx is periodic with period ____________.
The period of the function f(x) = tan3x is ____________.
