English

If Sin − 1 ( 2 a 1 − a 2 ) + Cos − 1 ( 1 − a 2 1 + a 2 ) = Tan − 1 ( 2 X 1 − X 2 ) , Where a , X ∈ ( 0 , 1 ) , Then, the Value of X is (A) 0 (B) a 2 (C) a (D) 2 a 1 − a 2 - Mathematics

Advertisements
Advertisements

Question

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

 

Options

  • 0

  • `a/2`

  •  a

  • `(2a)/(1-a^2)`

MCQ
Advertisements

Solution

\[\sin^{- 1} \left( \frac{2a}{1 - a^2} \right) + \cos^{- 1} \left( \frac{1 - a^2}{1 + a^2} \right) = \tan^{- 1} \left( \frac{2x}{1 - x^2} \right)\]
\[ \Rightarrow 2 \tan^{- 1} a + 2 \tan^{- 1} a = 2 \tan^{- 1} x\]
\[ \Rightarrow 4 \tan^{- 1} a = 2 \tan^{- 1} x\]
\[ \Rightarrow 2 \tan^{- 1} a = \tan^{- 1} x\]
\[ \Rightarrow \tan^{- 1} \left( \frac{2a}{1 - a^2} \right) = \tan^{- 1} x\]
\[ \Rightarrow x = \frac{2a}{1 - a^2}\]

Hence, the correct answer is option(d).

shaalaa.com
  Is there an error in this question or solution?
Chapter 4: Inverse Trigonometric Functions - Exercise 4.16 [Page 122]

APPEARS IN

RD Sharma Mathematics [English] Class 12
Chapter 4 Inverse Trigonometric Functions
Exercise 4.16 | Q 31 | Page 122

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`


 

Show that:

`2 sin^-1 (3/5)-tan^-1 (17/31)=pi/4`

 

 

Find the domain of `f(x)=cos^-1x+cosx.`


​Find the principal values of the following:

`cos^-1(-1/sqrt2)`


Evaluate the following:

`cos^-1(cos4)`


Evaluate the following:

`tan^-1(tan  (6pi)/7)`


Evaluate the following:

`tan^-1(tan  (7pi)/6)`


Evaluate the following:

`sec^-1(sec  (13pi)/4)`


Write the following in the simplest form:

`tan^-1{(sqrt(1+x^2)-1)/x},x !=0`


Evaluate the following:

`cot(cos^-1  3/5)`


Prove the following result

`sin(cos^-1  3/5+sin^-1  5/13)=63/65`


Evaluate:

`sec{cot^-1(-5/12)}`


Evaluate:

`tan{cos^-1(-7/25)}`


Evaluate: 

`cot(sin^-1  3/4+sec^-1  4/3)`


Evaluate:

`sin(tan^-1x+tan^-1  1/x)` for x > 0


Evaluate:

`cot(tan^-1a+cot^-1a)`


Prove the following result:

`sin^-1  12/13+cos^-1  4/5+tan^-1  63/16=pi`


Solve the following equation for x:

tan−1(x −1) + tan−1x tan−1(x + 1) = tan−13x


Solve the following equation for x:

`tan^-1  x/2+tan^-1  x/3=pi/4, 0<x<sqrt6`


Evaluate the following:

`sin(1/2cos^-1  4/5)`


`2sin^-1  3/5-tan^-1  17/31=pi/4`


Prove that

`tan^-1((1-x^2)/(2x))+cot^-1((1-x^2)/(2x))=pi/2`


For any a, b, x, y > 0, prove that:

`2/3tan^-1((3ab^2-a^3)/(b^3-3a^2b))+2/3tan^-1((3xy^2-x^3)/(y^3-3x^2y))=tan^-1  (2alphabeta)/(alpha^2-beta^2)`

`where  alpha =-ax+by, beta=bx+ay`


If `sin^-1x+sin^-1y+sin^-1z=(3pi)/2,`  then write the value of x + y + z.


Write the value of sin (cot−1 x).


Write the value of cos\[\left( 2 \sin^{- 1} \frac{1}{3} \right)\]


If tan−1 x + tan−1 y = `pi/4`,  then write the value of x + y + xy.


Write the value of sin−1 \[\left( \cos\frac{\pi}{9} \right)\]


If x < 0, y < 0 such that xy = 1, then write the value of tan1 x + tan−1 y.


Write the principal value of \[\tan^{- 1} 1 + \cos^{- 1} \left( - \frac{1}{2} \right)\]


Write the value of \[\cos^{- 1} \left( \cos\frac{14\pi}{3} \right)\]


Write the value of  `cot^-1(-x)`  for all `x in R` in terms of `cot^-1(x)`


2 tan−1 {cosec (tan−1 x) − tan (cot1 x)} is equal to


If  \[\cos^{- 1} \frac{x}{a} + \cos^{- 1} \frac{y}{b} = \alpha, then\frac{x^2}{a^2} - \frac{2xy}{ab}\cos \alpha + \frac{y^2}{b^2} = \]


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 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

 


Find the domain of `sec^(-1)(3x-1)`.


The value of sin `["cos"^-1 (7/25)]` is ____________.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×