हिंदी

If U = Cot − 1 √ Tan θ − Tan − 1 √ Tan θ Then , Tan ( π 4 − U 2 ) = (A) √ Tan θ (B) √ Cot θ (C) Tan θ (D) Cot θ - Mathematics

Advertisements
Advertisements

प्रश्न

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

विकल्प

  • `sqrt(tantheta`

  • `sqrt(cottheta)`

  •  tan θ

  • cot θ

MCQ
Advertisements

उत्तर

(a) `sqrt(tantheta`
Let \[y = \sqrt{\tan\theta}\]
Then, 
\[u = \cot^{- 1} \sqrt{\tan\theta} - \tan^{- 1} \sqrt{\tan\theta}\]
\[ \Rightarrow u = \cot^{- 1} y - \tan^{- 1} y\]
\[ \Rightarrow u = \frac{\pi}{2} - 2 \tan^{- 1} y \left[ \because \tan^{- 1} x + \cot^{- 1} x = \frac{\pi}{2} \right]\]
\[ \Rightarrow 2 \tan^{- 1} y = \frac{\pi}{2} - u \]
\[ \Rightarrow \tan^{- 1} y = \frac{\pi}{4} - \frac{u}{2}\]
\[ \Rightarrow y = \tan\left( \frac{\pi}{4} - \frac{u}{2} \right)\]
\[ \Rightarrow \sqrt{\tan\theta} = \tan\left( \frac{\pi}{4} - \frac{u}{2} \right) \left[ \because y = \sqrt{\tan\theta} \right]\]
\[\]

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 4: Inverse Trigonometric Functions - Exercise 4.16 [पृष्ठ १२०]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 12
अध्याय 4 Inverse Trigonometric Functions
Exercise 4.16 | Q 12 | पृष्ठ १२०

वीडियो ट्यूटोरियलVIEW ALL [2]

संबंधित प्रश्न

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


If sin [cot−1 (x+1)] = cos(tan1x), then find x.


 

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

 

Evaluate the following:

`cos^-1{cos  (5pi)/4}`


Evaluate the following:

`cos^-1{cos  ((4pi)/3)}`


Evaluate the following:

`cos^-1(cos4)`


Evaluate the following:

`tan^-1(tan  (7pi)/6)`


Evaluate the following:

`\text(cosec)^-1(\text{cosec}  pi/4)`


Evaluate the following:

`cot^-1(cot  (4pi)/3)`


Evaluate the following:

`cot^-1(cot  (9pi)/4)`


Write the following in the simplest form:

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


Write the following in the simplest form:

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


Write the following in the simplest form:

`tan^-1sqrt((a-x)/(a+x)),-a<x<a`


Write the following in the simplest form:

`sin^-1{(sqrt(1+x)+sqrt(1-x))/2},0<x<1`


Prove the following result

`tan(cos^-1  4/5+tan^-1  2/3)=17/6`


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


Solve the following equation for x:

 cot−1x − cot−1(x + 2) =`pi/12`, > 0


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


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


Prove that: `cos^-1  4/5+cos^-1  12/13=cos^-1  33/65`


Evaluate the following:

`sin(2tan^-1  2/3)+cos(tan^-1sqrt3)`


`sin^-1  4/5+2tan^-1  1/3=pi/2`


`2tan^-1(1/2)+tan^-1(1/7)=tan^-1(31/17)`


`4tan^-1  1/5-tan^-1  1/239=pi/4`


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`


If x > 1, then write the value of sin−1 `((2x)/(1+x^2))` in terms of tan−1 x.


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


Evaluate sin

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


Write the value of cos\[\left( \frac{1}{2} \cos^{- 1} \frac{3}{5} \right)\]


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


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


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


\[\text{ If }\cos^{- 1} \frac{x}{3} + \cos^{- 1} \frac{y}{2} = \frac{\theta}{2}, \text{ then }4 x^2 - 12xy \cos\frac{\theta}{2} + 9 y^2 =\]


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

 


If tan−1 (cot θ) = 2 θ, then θ =

 


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

 


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

 


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×