If tan - 1x - 1x - 2 - tan - 1x - 1x - 2 - n4- Find x

If tan^ 1x 1x 2+tan^ 1x+1x+2=π4 We know than tan^ 1x+tan^ 1y=tan^ 1(x+y)(1 xy) tan^ 1(x 1x 2)+tan^ 1x+1x+2=tan^ 1[x 1x 2+x+1x+21 x 1x 2.x+1x+2] ...

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Byju's Answer

Standard XII

Mathematics

Domain and Range of Basic Inverse Trigonometric Functions

If tan - 1x...

Question

If ${tan}^{- 1} (\frac{x - 1}{x - 2}) + {tan}^{- 1} (\frac{x + 1}{x + 2}) = \frac{n}{4}$ . Find $x$

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Solution

If ${tan}^{- 1} \frac{x - 1}{x - 2} + {tan}^{- 1} \frac{x + 1}{x + 2} = \frac{π}{4}$

We know than
${tan}^{- 1} x + {tan}^{- 1} y = {tan}^{- 1} \frac{(x + y)}{(1 - x y)}$

${tan}^{- 1} (\frac{x - 1}{x - 2}) + {tan}^{- 1} \frac{x + 1}{x + 2} = {tan}^{- 1} ⎡ ⎢ ⎢ ⎢ ⎣ \frac{\frac{x - 1}{x - 2} + \frac{x + 1}{x + 2}}{1 - \frac{x - 1}{x - 2} . \frac{x + 1}{x + 2}} ⎤ ⎥ ⎥ ⎥ ⎦$

$= {tan}^{- 1} ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ \frac{\frac{(x - 1) (x + 2) + (x + 1) (x - 2)}{(x - 2) (x + 2)}}{\frac{(x - 2) (x + 2) - (x - 1) (x + 2)}{(x - 2) (x + 2)}} ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦$

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

$= {tan}^{- 1} [\frac{x^{2} + 2 x - x - 2 + x^{2} - 2 x + x - 2}{x^{2} - x^{2} - 4 + 1}]$

$= {tan}^{- 1} [\frac{2 x^{2} - y}{- 3}]$

$∴ {tan}^{- 1} [\frac{2 x^{2} - y}{- 3}] = \frac{π}{4}$

$\frac{2 x^{2} - y}{- 3} = tan \frac{π}{4}$

$\frac{2 x^{2} - 4}{- 3} = 1$

$2 x^{2} - 4 = - 3$

$2 x^{2} = 4 - 3 = 1$

$x^{2} = \frac{1}{2}$

$x = \pm \frac{1}{\sqrt{2}}$

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Similar questions

Q. If

{tan}^{- 1} (\frac{x - 1}{x - 2}) + {tan}^{- 1} (\frac{x + 1}{x + 2}) = \frac{π}{4}

, find the value of

x

Prove that: $t a n^{- 1} (\frac{\sqrt{1 + x} - \sqrt{1 - x}}{\sqrt{1 + x} + \sqrt{1 - x}}) = \frac{π}{4} - \frac{1}{2} c o s^{- 1} x; - \frac{1}{\sqrt{2}} \leq x \leq 1$ .

OR If $t a n^{- 1} (\frac{x - 2}{x - 4}) + t a n^{- 1} (\frac{x + 2}{x + 4}) = \frac{π}{4}$ , find the value of x.

Q. Solve the following equations for x:
(i) tan⁻¹2x + tan⁻¹3x = nπ +

\frac{3 π}{4}

(ii) tan⁻¹(x + 1) + tan⁻¹(x − 1) = tan⁻¹

\frac{8}{31}

(iii) tan⁻¹(x −1) + tan⁻¹x tan⁻¹(x + 1) = tan⁻¹3x
(iv) tan⁻¹

(\frac{1 - x}{1 + x}) - \frac{1}{2}

tan⁻¹x = 0, where x > 0
(v) cot⁻¹x − cot⁻¹(x + 2) =

\frac{π}{12}

, x > 0
(vi) tan⁻¹(x + 2) + tan⁻¹(x − 2) = tan⁻¹

(\frac{8}{79})

, x > 0
(vii)

\tan^{- 1} \frac{x}{2} + \tan^{- 1} \frac{x}{3} = \frac{π}{4}, 0 < x < \sqrt{6}

(viii)

\tan^{- 1} (\frac{x - 2}{x - 4}) + \tan^{- 1} (\frac{x + 2}{x + 4}) = \frac{π}{4}

(ix)

\tan^{- 1} (2 + x) + \tan^{- 1} (2 - x) = \tan^{- 1} \frac{2}{3}, where x < - \sqrt{3} or, x > \sqrt{3}

(x)

\tan^{- 1} \frac{x - 2}{x - 1} + \tan^{- 1} \frac{x + 2}{x + 1} = \frac{π}{4}

Q. Solve :

{tan}^{- 1} (\frac{x - 1}{x - 2}) + {tan}^{- 1} (\frac{x + 1}{x + 2}) = \frac{π}{4}

Q. Solve for x:

{tan}^{- 1} (\frac{x - 1}{x - 2}) + {tan}^{- 1} (\frac{x + 1}{x + 2}) = \frac{π}{4}

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If tan−1(x−1x−2)+tan−1(x+1x+2)=n4. Find x

If ${tan}^{- 1} (\frac{x - 1}{x - 2}) + {tan}^{- 1} (\frac{x + 1}{x + 2}) = \frac{n}{4}$ . Find $x$