मराठी

If Cos − 1 X > Sin − 1 X , Then (A) 1 √ 2 < X ≤ 1 (B) 0 ≤ X < 1 √ 2 (C) − 1 ≤ X < 1 √ 2 (D) X > 0 - Mathematics

Advertisements
Advertisements

प्रश्न

If \[\cos^{- 1} x > \sin^{- 1} x\], then

पर्याय

  • \[\frac{1}{\sqrt{2}} < x \leq 1\]

  •  \[0 \leq x < \frac{1}{\sqrt{2}}\]

  •  \[- 1 \leq x < \frac{1}{\sqrt{2}}\]

  •  x > 0

MCQ
Advertisements

उत्तर

\[\cos^{- 1} x > \sin^{- 1} x\]
\[ \Rightarrow \cos^{- 1} x > \frac{\pi}{2} - \cos^{- 1} x\]
\[ \Rightarrow 2 \cos^{- 1} x > \frac{\pi}{2}\]
\[ \Rightarrow \cos^{- 1} x > \frac{\pi}{4}\]
\[ \Rightarrow x > \cos\frac{\pi}{4}\]
\[ \Rightarrow x > \frac{1}{\sqrt{2}}\]

We know that the maximum value of cosine fuction is 1.

\[\therefore \frac{1}{\sqrt{2}} < x \leq 1\]

Hence, the correct answer is option(a).

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 4: Inverse Trigonometric Functions - Exercise 4.16 [पृष्ठ १२१]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 12
पाठ 4 Inverse Trigonometric Functions
Exercise 4.16 | Q 26 | पृष्ठ १२१

व्हिडिओ ट्यूटोरियलVIEW ALL [2]

संबंधित प्रश्‍न

Write the value of `tan(2tan^(-1)(1/5))`


`sin^-1(sin3)`


`sin^-1(sin12)`


Evaluate the following:

`cos^-1(cos12)`


Evaluate the following:

`tan^-1(tan  (7pi)/6)`


Evaluate the following:

`sec^-1(sec  (9pi)/5)`


Evaluate the following:

`sec^-1{sec  (-(7pi)/3)}`


Evaluate the following:

`sec^-1(sec  (25pi)/6)`


Evaluate the following:

`cot^-1(cot  (19pi)/6)`


Write the following in the simplest form:

`tan^-1{sqrt(1+x^2)-x},x in R`


Write the following in the simplest form:

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


`5tan^-1x+3cot^-1x=2x`


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


If `cos^-1  x/2+cos^-1  y/3=alpha,` then prove that  `9x^2-12xy cosa+4y^2=36sin^2a.`


Prove that:

`2sin^-1  3/5=tan^-1  24/7`


If `sin^-1  (2a)/(1+a^2)-cos^-1  (1-b^2)/(1+b^2)=tan^-1  (2x)/(1-x^2)`, then prove that `x=(a-b)/(1+ab)`


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`


Write the range of tan−1 x.


Evaluate sin \[\left( \tan^{- 1} \frac{3}{4} \right)\]


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


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


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


Write the value of \[\sin^{- 1} \left( \sin\frac{3\pi}{5} \right)\]


Wnte the value of the expression \[\tan\left( \frac{\sin^{- 1} x + \cos^{- 1} x}{2} \right), \text { when } x = \frac{\sqrt{3}}{2}\]


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


If \[\cos\left( \tan^{- 1} x + \cot^{- 1} \sqrt{3} \right) = 0\] , find the value of x.

 

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 value of tan \[\left\{ \cos^{- 1} \frac{1}{5\sqrt{2}} - \sin^{- 1} \frac{4}{\sqrt{17}} \right\}\] is

 


If sin−1 − cos−1 x = `pi/6` , then x = 


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

 


The value of sin \[\left( \frac{1}{4} \sin^{- 1} \frac{\sqrt{63}}{8} \right)\] is

 


The value of \[\tan\left( \cos^{- 1} \frac{3}{5} + \tan^{- 1} \frac{1}{4} \right)\]

 


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.


Find the real solutions of the equation
`tan^-1 sqrt(x(x + 1)) + sin^-1 sqrt(x^2 + x + 1) = pi/2`


Find the simplified form of `cos^-1 (3/5 cosx + 4/5 sin x)`, where x ∈ `[(-3pi)/4, pi/4]`


Solve for x : {xcos(cot-1 x) + sin(cot-1 x)}= `51/50`


The value of tan `("cos"^-1  4/5 + "tan"^-1  2/3) =`


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×