Advertisements
Advertisements
प्रश्न
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
पर्याय
0
`a/2`
a
`(2a)/(1-a^2)`
Advertisements
उत्तर
\[\sin^{- 1} \left( \frac{2a}{1 - a^2} \right) + \cos^{- 1} \left( \frac{1 - a^2}{1 + a^2} \right) = \tan^{- 1} \left( \frac{2x}{1 - x^2} \right)\]
\[ \Rightarrow 2 \tan^{- 1} a + 2 \tan^{- 1} a = 2 \tan^{- 1} x\]
\[ \Rightarrow 4 \tan^{- 1} a = 2 \tan^{- 1} x\]
\[ \Rightarrow 2 \tan^{- 1} a = \tan^{- 1} x\]
\[ \Rightarrow \tan^{- 1} \left( \frac{2a}{1 - a^2} \right) = \tan^{- 1} x\]
\[ \Rightarrow x = \frac{2a}{1 - a^2}\]
Hence, the correct answer is option(d).
APPEARS IN
संबंधित प्रश्न
Prove that :
`2 tan^-1 (sqrt((a-b)/(a+b))tan(x/2))=cos^-1 ((a cos x+b)/(a+b cosx))`
Solve for x:
`2tan^(-1)(cosx)=tan^(-1)(2"cosec" x)`
If `tan^(-1)((x-2)/(x-4)) +tan^(-1)((x+2)/(x+4))=pi/4` ,find the value of x
`sin^-1(sin12)`
Evaluate the following:
`cos^-1{cos (5pi)/4}`
Evaluate the following:
`cos^-1{cos ((4pi)/3)}`
Evaluate the following:
`sec^-1(sec pi/3)`
Evaluate the following:
`sec^-1(sec (2pi)/3)`
Evaluate the following:
`sec^-1(sec (13pi)/4)`
Evaluate the following:
`sec^-1(sec (25pi)/6)`
Evaluate the following:
`cosec^-1(cosec (3pi)/4)`
Evaluate the following:
`cosec^-1(cosec (13pi)/6)`
Evaluate the following:
`cosec^-1{cosec (-(9pi)/4)}`
Evaluate the following:
`cot^-1(cot (4pi)/3)`
Evaluate the following:
`cot^-1{cot (-(8pi)/3)}`
Write the following in the simplest form:
`tan^-1{(sqrt(1+x^2)-1)/x},x !=0`
Evaluate the following:
`sec(sin^-1 12/13)`
Evaluate the following:
`cot(cos^-1 3/5)`
Evaluate:
`cos(tan^-1 3/4)`
Evaluate: `sin{cos^-1(-3/5)+cot^-1(-5/12)}`
Evaluate:
`sin(tan^-1x+tan^-1 1/x)` for x < 0
If `cot(cos^-1 3/5+sin^-1x)=0`, find the values of x.
`5tan^-1x+3cot^-1x=2x`
Solve the following equation for x:
`tan^-1 (x-2)/(x-1)+tan^-1 (x+2)/(x+1)=pi/4`
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))`
`tan^-1 1/7+2tan^-1 1/3=pi/4`
`2sin^-1 3/5-tan^-1 17/31=pi/4`
If x < 0, then write the value of cos−1 `((1-x^2)/(1+x^2))` in terms of tan−1 x.
2 tan−1 {cosec (tan−1 x) − tan (cot−1 x)} is equal to
The positive integral solution of the equation
\[\tan^{- 1} x + \cos^{- 1} \frac{y}{\sqrt{1 + y^2}} = \sin^{- 1} \frac{3}{\sqrt{10}}\text{ is }\]
\[\text{ If } u = \cot^{- 1} \sqrt{\tan \theta} - \tan^{- 1} \sqrt{\tan \theta}\text{ then }, \tan\left( \frac{\pi}{4} - \frac{u}{2} \right) =\]
\[\text{ If }\cos^{- 1} \frac{x}{3} + \cos^{- 1} \frac{y}{2} = \frac{\theta}{2}, \text{ then }4 x^2 - 12xy \cos\frac{\theta}{2} + 9 y^2 =\]
Let f (x) = `e^(cos^-1){sin(x+pi/3}.`
Then, f (8π/9) =
The value of \[\cos^{- 1} \left( \cos\frac{5\pi}{3} \right) + \sin^{- 1} \left( \sin\frac{5\pi}{3} \right)\] is
sin \[\left\{ 2 \cos^{- 1} \left( \frac{- 3}{5} \right) \right\}\] is equal to
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
The domain of \[\cos^{- 1} \left( x^2 - 4 \right)\] is
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.
