हिंदी

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

Advertisements
Advertisements

प्रश्न

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

Advertisements

उत्तर

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

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 4: Inverse Trigonometric Functions - Exercise 4.15 [पृष्ठ ११८]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 12
अध्याय 4 Inverse Trigonometric Functions
Exercise 4.15 | Q 50 | पृष्ठ ११८

वीडियो ट्यूटोरियलVIEW ALL [2]

संबंधित प्रश्न

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`


 

Prove that :

`2 tan^-1 (sqrt((a-b)/(a+b))tan(x/2))=cos^-1 ((a cos x+b)/(a+b cosx))`

 

Solve the following for x:

`sin^(-1)(1-x)-2sin^-1 x=pi/2`


 

If `tan^(-1)((x-2)/(x-4)) +tan^(-1)((x+2)/(x+4))=pi/4` ,find the value of x

 

`sin^-1(sin  (17pi)/8)`


`sin^-1(sin2)`


Evaluate the following:

`cos^-1{cos  ((4pi)/3)}`


Evaluate the following:

`tan^-1(tan  (7pi)/6)`


Evaluate the following:

`sec^-1(sec  (9pi)/5)`


Evaluate the following:

`sec^-1(sec  (13pi)/4)`


Evaluate the following:

`cosec^-1(cosec  (6pi)/5)`


Write the following in the simplest form:

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


Evaluate the following:

`sin(sec^-1  17/8)`


Evaluate the following:

`cos(tan^-1  24/7)`


Prove the following result

`tan(cos^-1  4/5+tan^-1  2/3)=17/6`


Prove the following result

`sin(cos^-1  3/5+sin^-1  5/13)=63/65`


Find the value of `tan^-1  (x/y)-tan^-1((x-y)/(x+y))`


Solve the following equation for x:

`tan^-1(2+x)+tan^-1(2-x)=tan^-1  2/3, where  x< -sqrt3 or, x>sqrt3`


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


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


`4tan^-1  1/5-tan^-1  1/239=pi/4`


Prove that

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


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:

`tan^-1((2x)/(1-x^2))+cot^-1((1-x^2)/(2x))=(2pi)/3,x>0`


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 value of cos−1 (cos 1540°).


Write the value of sin−1

\[\left( \sin( -{600}°) \right)\].

 

 


Write the value of sin1 (sin 1550°).


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


Write the value of cos−1 (cos 6).


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


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


If \[\sin^{- 1} \left( \frac{1}{3} \right) + \cos^{- 1} x = \frac{\pi}{2},\] then find x.

 


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


The value of \[\sin^{- 1} \left( \cos\frac{33\pi}{5} \right)\] is 

 


If x = a (2θ – sin 2θ) and y = a (1 – cos 2θ), find \[\frac{dy}{dx}\] When  \[\theta = \frac{\pi}{3}\] .


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


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×