English

Wnte the Value of the Expression Tan ( Sin − 1 X + Cos − 1 X 2 ) , When X = √ 3 2 - Mathematics

Advertisements
Advertisements

Question

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

Solution

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

shaalaa.com
  Is there an error in this question or solution?
Chapter 4: Inverse Trigonometric Functions - Exercise 4.15 [Page 118]

APPEARS IN

RD Sharma Mathematics [English] Class 12
Chapter 4 Inverse Trigonometric Functions
Exercise 4.15 | Q 50 | Page 118

RELATED QUESTIONS

 

Show that:

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

 

 

If `(sin^-1x)^2 + (sin^-1y)^2+(sin^-1z)^2=3/4pi^2,`  find the value of x2 + y2 + z2 


Evaluate the following:

`tan^-1(tan1)`


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  pi/3)`


Write the following in the simplest form:

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


Evaluate the following:

`tan(cos^-1  8/17)`


Evaluate the following:

`cos(tan^-1  24/7)`


Solve: `cos(sin^-1x)=1/6`


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


Evaluate: 

`cot(sin^-1  3/4+sec^-1  4/3)`


If `sin^-1x+sin^-1y=pi/3`  and  `cos^-1x-cos^-1y=pi/6`,  find the values of x and y.


Solve the following equation for x:

`tan^-1  2x+tan^-1  3x = npi+(3pi)/4`


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:

 cot−1x − cot−1(x + 2) =`pi/12`, > 0


Evaluate: `cos(sin^-1  3/5+sin^-1  5/13)`


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


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)`


Show that `2tan^-1x+sin^-1  (2x)/(1+x^2)` is constant for x ≥ 1, find that constant.


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`


If x < 0, then write the value of cos−1 `((1-x^2)/(1+x^2))` in terms of tan−1 x.


Write the value of tan1 x + tan−1 `(1/x)`  for x < 0.


Evaluate sin

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


Show that \[\sin^{- 1} (2x\sqrt{1 - x^2}) = 2 \sin^{- 1} x\]


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


Write the value of \[\cos\left( \sin^{- 1} x + \cos^{- 1} x \right), \left| x \right| \leq 1\]


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


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


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


The value of  \[\sin\left( 2\left( \tan^{- 1} 0 . 75 \right) \right)\] is equal to

 


The domain of  \[\cos^{- 1} \left( x^2 - 4 \right)\] is

 


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

 


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


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


Find the value of x, if tan `[sec^(-1) (1/x) ] = sin ( tan^(-1) 2) , x > 0 `.


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


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×