मराठी

If Tan − 1 ( √ 1 + X 2 − √ 1 − X 2 √ 1 + X 2 + √ 1 − X 2 ) = α, Then X2 = (A) Sin 2 α (B) Sin α (C) Cos 2 α (D) Cos α - Mathematics

Advertisements
Advertisements

प्रश्न

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 =



पर्याय

  • sin 2 α

  • sin α

  • cos 2 α

  • cos α

MCQ
Advertisements

उत्तर

(a) sin 2α
\[\tan^{- 1} \left( \frac{\sqrt{1 + x^2} - \sqrt{1 - x^2}}{\sqrt{1 + x^2} + \sqrt{1 - x^2}} \right) = \alpha\]
\[ \Rightarrow \frac{\sqrt{1 + x^2} - \sqrt{1 - x^2}}{\sqrt{1 + x^2} + \sqrt{1 - x^2}} = \tan\alpha\]
\[\]
\[ \Rightarrow \frac{\sqrt{1 + x^2} - \sqrt{1 - x^2}}{\sqrt{1 + x^2} + \sqrt{1 - x^2}} \times \frac{\sqrt{1 + x^2} - \sqrt{1 - x^2}}{\sqrt{1 + x^2} - \sqrt{1 - x^2}} = \tan\alpha\]
\[ \Rightarrow \frac{\left( \sqrt{1 + x^2} \right)^2 + \left( \sqrt{1 - x^2} \right)^2 - 2\sqrt{1 + x^2}\sqrt{1 - x^2}}{\left( \sqrt{1 + x^2} \right)^2 - \left( \sqrt{1 - x^2} \right)^2} = \tan\alpha\]
\[ \Rightarrow \frac{1 - \sqrt{1 - x^4}}{x^2} = \tan\alpha\]
\[ \Rightarrow x^2 \tan\alpha = 1 - \sqrt{1 - x^4}\]
\[ \Rightarrow \sqrt{1 - x^4} = 1 - x^2 \tan\alpha\]
\[ \Rightarrow 1 - x^4 = 1 + x^4 \tan^2 \alpha - 2 x^2 \tan\alpha\]
\[ \Rightarrow x^4 + x^4 \tan^2 \alpha - 2 x^2 \tan\alpha = 0\]
\[ \Rightarrow x^4 \sec^2 \alpha - 2 x^2 \tan\alpha = 0\]
\[ \Rightarrow x^2 \left( x^2 \sec^2 \alpha - 2\tan\alpha \right) = 0\]
\[ \Rightarrow x^2 \sec^2 \alpha - 2\tan\alpha = 0 \left[ \because x^2 \neq 0 \right]\]
\[ \Rightarrow x^2 \sec^2 \alpha = 2\tan\alpha\]
\[ \Rightarrow x^2 = \frac{2\tan\alpha}{\sec^2 \alpha} = 2\sin\alpha\cos\alpha = \sin2\alpha\]

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 4: Inverse Trigonometric Functions - Exercise 4.16 [पृष्ठ ११९]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 12
पाठ 4 Inverse Trigonometric Functions
Exercise 4.16 | Q 1 | पृष्ठ ११९

व्हिडिओ ट्यूटोरियलVIEW ALL [2]

संबंधित प्रश्‍न

Solve the equation for x:sin1x+sin1(1x)=cos1x


Solve the following for x :

`tan^(-1)((x-2)/(x-3))+tan^(-1)((x+2)/(x+3))=pi/4,|x|<1`


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


​Find the principal values of the following:
`cos^-1(-sqrt3/2)`


`sin^-1(sin  (7pi)/6)`


Evaluate the following:

`tan^-1(tan  pi/3)`


Evaluate the following:

`tan^-1(tan4)`


Evaluate the following:

`sec^-1(sec  (5pi)/4)`


Evaluate the following:

`sec^-1(sec  (13pi)/4)`


Evaluate the following:

`cosec^-1(cosec  (13pi)/6)`


Evaluate the following:

`cot^-1(cot  (9pi)/4)`


Evaluate the following:

`cot^-1{cot  ((21pi)/4)}`


Write the following in the simplest form:

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


Evaluate the following:

`sin(cos^-1  5/13)`


Evaluate the following:

`sin(tan^-1  24/7)`


Evaluate the following:

`cosec(cos^-1  3/5)`


Evaluate: `sin{cos^-1(-3/5)+cot^-1(-5/12)}`


Evaluate:

`sin(tan^-1x+tan^-1  1/x)` for x > 0


Solve the following equation for x:

`tan^-1((1-x)/(1+x))-1/2 tan^-1x` = 0, where x > 0


Solve the following equation for x:

tan−1(x + 2) + tan−1(x − 2) = tan−1 `(8/79)`, x > 0


Solve the following equation for x:

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


Solve the equation `cos^-1  a/x-cos^-1  b/x=cos^-1  1/b-cos^-1  1/a`


Solve `cos^-1sqrt3x+cos^-1x=pi/2`


`tan^-1  1/4+tan^-1  2/9=1/2cos^-1  3/2=1/2sin^-1(4/5)`


`tan^-1  1/7+2tan^-1  1/3=pi/4`


Prove that

`sin{tan^-1  (1-x^2)/(2x)+cos^-1  (1-x^2)/(2x)}=1`


Solve the following equation for x:

`tan^-1  1/4+2tan^-1  1/5+tan^-1  1/6+tan^-1  1/x=pi/4`


Solve the following equation for x:

`3sin^-1  (2x)/(1+x^2)-4cos^-1  (1-x^2)/(1+x^2)+2tan^-1  (2x)/(1-x^2)=pi/3`


Write the difference between maximum and minimum values of  sin−1 x for x ∈ [− 1, 1].


What is the value of cos−1 `(cos  (2x)/3)+sin^-1(sin  (2x)/3)?`


Write the value of cos−1 (cos 1540°).


Evaluate sin

\[\left( \frac{1}{2} \cos^{- 1} \frac{4}{5} \right)\]


Write the value of cos−1 \[\left( \tan\frac{3\pi}{4} \right)\]


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


If α = \[\tan^{- 1} \left( \frac{\sqrt{3}x}{2y - x} \right), \beta = \tan^{- 1} \left( \frac{2x - y}{\sqrt{3}y} \right),\] 
 then α − β =


Let f (x) = `e^(cos^-1){sin(x+pi/3}.`
Then, f (8π/9) = 


It \[\tan^{- 1} \frac{x + 1}{x - 1} + \tan^{- 1} \frac{x - 1}{x} = \tan^{- 1}\]   (−7), then the value of x is

 


If tan−1 (cot θ) = 2 θ, then θ =

 


Find : \[\int\frac{2 \cos x}{\left( 1 - \sin x \right) \left( 1 + \sin^2 x \right)}dx\] .


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.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×