मराठी

Solve for X : Tan − 1 ( X − 2 X − 1 ) + Tan − 1 ( X + 2 X + 1 ) = π 4 . - Mathematics

Advertisements
Advertisements

प्रश्न

Solve for x : \[\tan^{- 1} \left( \frac{x - 2}{x - 1} \right) + \tan^{- 1} \left( \frac{x + 2}{x + 1} \right) = \frac{\pi}{4}\] .

Advertisements

उत्तर

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

\[ \Rightarrow \tan^{- 1} \left( \frac{x - 2}{x - 1} \right) + \tan^{- 1} \left( \frac{x + 2}{x + 1} \right) = \tan^{- 1} 1\]

\[ \Rightarrow \tan^{- 1} \left( \frac{x - 2}{x - 1} \right) = \tan^{- 1} 1 - \tan^{- 1} \left( \frac{x + 2}{x + 1} \right)\]

\[ \Rightarrow \tan^{- 1} \left( \frac{x - 2}{x - 1} \right) = \tan^{- 1} \left( \frac{1 - \frac{x + 2}{x + 1}}{1 + \frac{x + 2}{x + 1}} \right)\]

\[ \Rightarrow \tan^{- 1} \left( \frac{x - 2}{x - 1} \right) = \tan^{- 1} \left( \frac{x + 1 - x - 2}{x + 1 + x + 2} \right)\]

\[ \Rightarrow \tan^{- 1} \left( \frac{x - 2}{x - 1} \right) = \tan^{- 1} \left( \frac{- 1}{2x + 3} \right)\]

\[ \Rightarrow \frac{x - 2}{x - 1} = \frac{- 1}{2x + 3}\]

\[ \Rightarrow 2 x^2 + 3x - 4x - 6 = - x + 1\]

\[ \Rightarrow 2 x^2 = 1 + 6\]

\[ \Rightarrow x^2 = 7\]

\[ \Rightarrow x = \pm \sqrt{\frac{7}{2}}\]

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
2015-2016 (March) Foreign Set 2

संबंधित प्रश्‍न

If a line makes angles 90°, 60° and θ with x, y and z-axis respectively, where θ is acute, then find θ.


If `tan^-1(2x)+tan^-1(3x)=pi/4`, then find the value of ‘x’.


Prove the following: 

3cos−1x = cos−1(4x3 − 3x), `x ∈ [1/2, 1]`


Find the value of `cot(tan^(-1) a + cot^(-1) a)`


Solve the following equation:

2 tan−1 (cos x) = tan−1 (2 cosec x)


Prove that `tan {pi/4 + 1/2 cos^(-1)  a/b} + tan {pi/4 - 1/2 cos^(-1)  a/b} = (2b)/a`


Find the value of `sin^-1[cos(sin^-1 (sqrt(3)/2))]`


Choose the correct alternative:

`sin^-1  3/5 - cos^-1  13/13 + sec^-1  5/3 - "cosec"^-1  13/12` is equal to


Choose the correct alternative:

`sin^-1 (tan  pi/4) - sin^-1 (sqrt(3/x)) = pi/6`. Then x is a root of the equation


Choose the correct alternative:

sin–1(2 cos2x – 1) + cos1(1 – 2 sin2x) =


Choose the correct alternative:

sin(tan–1x), |x| < 1 is equal to


Evaluate: `tan^-1 sqrt(3) - sec^-1(-2)`.


Evaluate `cos[sin^-1  1/4 + sec^-1  4/3]`


If α ≤ 2 sin–1x + cos–1x ≤ β, then ______.


If `sin^-1 ((2"a")/(1 + "a"^2)) + cos^-1 ((1 - "a"^2)/(1 + "a"^2)) = tan^-1 ((2x)/(1 - x^2))`. where a, x ∈ ] 0, 1, then the value of x is ______.


If y = `2 tan^-1x + sin^-1 ((2x)/(1 + x^2))` for all x, then ______ < y < ______.


The value of cot–1(–x) for all x ∈ R in terms of cot–1x is ______.


The value of cot `("cosec"^-1 5/3 + "tan"^-1 2/3)` is ____________.


If tan-1 2x + tan-1 3x = `pi/4,` then x is ____________.


sin (tan−1 x), where |x| < 1, is equal to:


The value of `"cos"^-1 ("cos" ((33pi)/5))` is ____________.


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


`"tan"^-1 1 + "cos"^-1 ((-1)/2) + "sin"^-1 ((-1)/2)`


`"cos"^-1 (1/2)`


If `3  "sin"^-1 ((2"x")/(1 + "x"^2)) - 4  "cos"^-1 ((1 - "x"^2)/(1 + "x"^2)) + 2 "tan"^-1 ((2"x")/(1 - "x"^2)) = pi/3` then x is equal to ____________.


Solve for x : `{"x cos" ("cot"^-1 "x") + "sin" ("cot"^-1 "x")}^2` = `51/50


What is the simplest form of `tan^-1  sqrt(1 - x^2 - 1)/x, x ≠ 0`


If `tan^-1 ((x - 1)/(x + 1)) + tan^-1 ((2x - 1)/(2x + 1)) = tan^-1 (23/36)` = then prove that 24x2 – 23x – 12 = 0


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×