Advertisements
Advertisements
प्रश्न
Show that \[\sin^{- 1} (2x\sqrt{1 - x^2}) = 2 \sin^{- 1} x\]
उत्तर
We have
\[LHS = \sin^{- 1} \left( 2x\sqrt{1 - x^2} \right)\]
\[\text{Putting }x = \sin a, \text{we get}\]
\[ = \sin^{- 1} \left( 2 \sin a\sqrt{1 - \sin^2 a} \right) \]
\[ = \sin^{- 1} \left( 2\sin a \cos a \right)\]
\[ = \sin^{- 1} \left( \sin 2a \right)\]
\[ = 2a\]
\[ = 2 \sin^{- 1} x \left( \because x = \sin a \right)\]
\[\]
APPEARS IN
संबंधित प्रश्न
Find the principal values of the following:
`cos^-1(-sqrt3/2)`
`sin^-1(sin2)`
Evaluate the following:
`cos^-1(cos3)`
Evaluate the following:
`tan^-1(tan (7pi)/6)`
Evaluate the following:
`tan^-1(tan (9pi)/4)`
Evaluate the following:
`tan^-1(tan12)`
Evaluate the following:
`cosec^-1(cosec (6pi)/5)`
Write the following in the simplest form:
`tan^-1(x/(a+sqrt(a^2-x^2))),-a<x<a`
Evaluate the following:
`cosec(cos^-1 3/5)`
Evaluate the following:
`cot(cos^-1 3/5)`
Evaluate:
`cot{sec^-1(-13/5)}`
Evaluate:
`sin(tan^-1x+tan^-1 1/x)` for x < 0
Solve the following equation for x:
tan−1(x + 1) + tan−1(x − 1) = tan−1`8/31`
Sum the following series:
`tan^-1 1/3+tan^-1 2/9+tan^-1 4/33+...+tan^-1 (2^(n-1))/(1+2^(2n-1))`
`sin^-1 63/65=sin^-1 5/13+cos^-1 3/5`
Evaluate the following:
`sin(2tan^-1 2/3)+cos(tan^-1sqrt3)`
`2sin^-1 3/5-tan^-1 17/31=pi/4`
`2tan^-1 1/5+tan^-1 1/8=tan^-1 4/7`
`2tan^-1 3/4-tan^-1 17/31=pi/4`
Write the value of tan−1x + tan−1 `(1/x)`for x > 0.
What is the value of cos−1 `(cos (2x)/3)+sin^-1(sin (2x)/3)?`
Write the value of tan−1\[\left\{ \tan\left( \frac{15\pi}{4} \right) \right\}\]
Evaluate: \[\sin^{- 1} \left( \sin\frac{3\pi}{5} \right)\]
Write the principal value of `sin^-1(-1/2)`
Write the principal value of \[\tan^{- 1} 1 + \cos^{- 1} \left( - \frac{1}{2} \right)\]
Write the principal value of \[\cos^{- 1} \left( \cos680^\circ \right)\]
Wnte the value of the expression \[\tan\left( \frac{\sin^{- 1} x + \cos^{- 1} x}{2} \right), \text { when } x = \frac{\sqrt{3}}{2}\]
Find the value of \[2 \sec^{- 1} 2 + \sin^{- 1} \left( \frac{1}{2} \right)\]
If sin−1 x − cos−1 x = `pi/6` , then x =
If α = \[\tan^{- 1} \left( \tan\frac{5\pi}{4} \right) \text{ and }\beta = \tan^{- 1} \left( - \tan\frac{2\pi}{3} \right)\] , then
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
If θ = sin−1 {sin (−600°)}, then one of the possible values of θ is
\[\cot\left( \frac{\pi}{4} - 2 \cot^{- 1} 3 \right) =\]
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
If x = a (2θ – sin 2θ) and y = a (1 – cos 2θ), find \[\frac{dy}{dx}\] When \[\theta = \frac{\pi}{3}\] .
Find the real solutions of the equation
`tan^-1 sqrt(x(x + 1)) + sin^-1 sqrt(x^2 + x + 1) = pi/2`
