मराठी

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

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)`

MCQ
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).

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 4: Inverse Trigonometric Functions - Exercise 4.16 [पृष्ठ १२२]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 12
पाठ 4 Inverse Trigonometric Functions
Exercise 4.16 | Q 31 | पृष्ठ १२२

व्हिडिओ ट्यूटोरियलVIEW ALL [2]

संबंधित प्रश्‍न

​Find the principal values of the following:

`cos^-1(sin   (4pi)/3)`


`sin^-1(sin3)`


`sin^-1(sin2)`


Evaluate the following:

`cos^-1{cos  (5pi)/4}`


Evaluate the following:

`cos^-1(cos5)`


Evaluate the following:

`tan^-1(tan  (6pi)/7)`


Evaluate the following:

`sec^-1(sec  (9pi)/5)`


Evaluate the following:

`cosec^-1(cosec  (11pi)/6)`


Evaluate the following:

`tan(cos^-1  8/17)`


Evaluate:

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


Evaluate:

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


Evaluate:

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


`tan^-1x+2cot^-1x=(2x)/3`


Solve the following equation for x:

`tan^-1((x-2)/(x-4))+tan^-1((x+2)/(x+4))=pi/4`


`(9pi)/8-9/4sin^-1  1/3=9/4sin^-1  (2sqrt2)/3`


Solve the following:

`cos^-1x+sin^-1  x/2=π/6`


Evaluate the following:

`tan{2tan^-1  1/5-pi/4}`


Solve the following equation for x:

`2tan^-1(sinx)=tan^-1(2sinx),x!=pi/2`


Write the value of tan1x + 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 sin \[\left\{ \frac{\pi}{3} - \sin^{- 1} \left( - \frac{1}{2} \right) \right\}\]


Write the value of tan1\[\left\{ \tan\left( \frac{15\pi}{4} \right) \right\}\]


Write the value of \[\tan^{- 1} \frac{a}{b} - \tan^{- 1} \left( \frac{a - b}{a + b} \right)\]


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


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


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


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


Find the value of \[\cos^{- 1} \left( \cos\frac{13\pi}{6} \right)\]


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 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

 


\[\text{ If } u = \cot^{- 1} \sqrt{\tan \theta} - \tan^{- 1} \sqrt{\tan \theta}\text{ then }, \tan\left( \frac{\pi}{4} - \frac{u}{2} \right) =\]


If θ = sin−1 {sin (−600°)}, then one of the possible values of θ is

 


In a ∆ ABC, if C is a right angle, then
\[\tan^{- 1} \left( \frac{a}{b + c} \right) + \tan^{- 1} \left( \frac{b}{c + a} \right) =\]

 

 


\[\cot\left( \frac{\pi}{4} - 2 \cot^{- 1} 3 \right) =\] 

 


If > 1, then \[2 \tan^{- 1} x + \sin^{- 1} \left( \frac{2x}{1 + x^2} \right)\] is equal to

 


If y = sin (sin x), prove that \[\frac{d^2 y}{d x^2} + \tan x \frac{dy}{dx} + y \cos^2 x = 0 .\]


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


The period of the function f(x) = tan3x is ____________.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×