हिंदी

Tan − 1 1 11 + Tan − 1 2 11 is Equal to (A) 0 (B) 1/2 (C) − 1 (D) None of These - Mathematics

Advertisements
Advertisements

प्रश्न

\[\tan^{- 1} \frac{1}{11} + \tan^{- 1} \frac{2}{11}\]  is equal to

 

 

विकल्प

  • 0

  • 1/2

  • − 1

  • none of these

MCQ
Advertisements

उत्तर

(d) none of these

We know that 
\[\tan^{- 1} x + \tan^{- 1} y = \tan^{- 1} \left( \frac{x + y}{1 - xy} \right)\]
Now,
\[\tan^{- 1} \frac{1}{11} + \tan^{- 1} \frac{2}{11} = \tan^{- 1} \left( \frac{\frac{1}{11} + \frac{2}{11}}{1 - \frac{1}{11}\frac{2}{11}} \right)\]
\[ = \tan^{- 1} \left( \frac{\frac{3}{11}}{\frac{121 - 2}{121}} \right)\]
\[ = \tan^{- 1} \left( \frac{\frac{3}{11}}{\frac{119}{121}} \right)\]
\[ = \tan^{- 1} \left( \frac{33}{119} \right)\]
\[ = 0 . 27\]

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 4: Inverse Trigonometric Functions - Exercise 4.16 [पृष्ठ १२१]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 12
अध्याय 4 Inverse Trigonometric Functions
Exercise 4.16 | Q 16 | पृष्ठ १२१

वीडियो ट्यूटोरियलVIEW ALL [2]

संबंधित प्रश्न

Solve for x:

`2tan^(-1)(cosx)=tan^(-1)(2"cosec" x)`


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


Find the domain of  `f(x) =2cos^-1 2x+sin^-1x.`


Evaluate the following:

`cos^-1{cos(-pi/4)}`


Evaluate the following:

`cos^-1{cos  (5pi)/4}`


Evaluate the following:

`cosec^-1(cosec  (6pi)/5)`


Evaluate the following:

`cosec^-1(cosec  (13pi)/6)`


Evaluate the following:

`cot^-1(cot  (4pi)/3)`


Write the following in the simplest form:

`cot^-1  a/sqrt(x^2-a^2),|  x  | > a`


Write the following in the simplest form:

`sin^-1{(x+sqrt(1-x^2))/sqrt2},-1<x<1`


Write the following in the simplest form:

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


Evaluate the following:

`cot(cos^-1  3/5)`


Prove the following result

`sin(cos^-1  3/5+sin^-1  5/13)=63/65`


Evaluate: 

`cot(sin^-1  3/4+sec^-1  4/3)`


Prove the following result:

`sin^-1  12/13+cos^-1  4/5+tan^-1  63/16=pi`


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)/(x-1)+tan^-1  (x+2)/(x+1)=pi/4`


Sum the following series:

`tan^-1  1/3+tan^-1  2/9+tan^-1  4/33+...+tan^-1  (2^(n-1))/(1+2^(2n-1))`


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


If `sin^-1x+sin^-1y+sin^-1z=(3pi)/2,`  then write the value of x + y + z.


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


Write the value of sin (cot−1 x).


Write the value of cos−1 (cos 1540°).


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


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


Write the value of \[\tan^{- 1} \frac{a}{b} - \tan^{- 1} \left( \frac{a - b}{a + b} \right)\]


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

 


Write the value of \[\cos^{- 1} \left( \cos\frac{14\pi}{3} \right)\]


The set of values of `\text(cosec)^-1(sqrt3/2)`


Write the value of  \[\tan^{- 1} \left( \frac{1}{x} \right)\]  for x < 0 in terms of `cot^-1x`


The number of real solutions of the equation \[\sqrt{1 + \cos 2x} = \sqrt{2} \sin^{- 1} (\sin x), - \pi \leq x \leq \pi\]


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


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

 


If tan−1 (cot θ) = 2 θ, then θ =

 


If \[\sin^{- 1} \left( \frac{2a}{1 - a^2} \right) + \cos^{- 1} \left( \frac{1 - a^2}{1 + a^2} \right) = \tan^{- 1} \left( \frac{2x}{1 - x^2} \right),\text{ where }a, x \in \left( 0, 1 \right)\] , then, the value of x is

 


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


Prove that : \[\cot^{- 1} \frac{\sqrt{1 + \sin x} + \sqrt{1 - \sin x}}{\sqrt{1 + \sin x} - \sqrt{1 - \sin x}} = \frac{x}{2}, 0 < x < \frac{\pi}{2}\] .


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


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


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×