If tan -1 x-1 - x-2 -tan -1 x-1 - x-2 - 4- then x-

The correct options are C #160; 1 /√( 2 ) #160; D 1 /√( 2 ) #160; We have, #160;tan ^ 1 x 1 / x 2 #160; +tan ^ 1 x+1 / x+2 #160; ...

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Byju's Answer

Standard XII

Mathematics

First Principle of Differentiation

If tan -1 x...

Question

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

\frac{1}{2}

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- \frac{1}{2}

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\frac{1}{\sqrt{2}}

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- \frac{1}{\sqrt{2}}

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Solution

The correct options are
C $\frac{1}{\sqrt{2}}$
D $- \frac{1}{\sqrt{2}}$
We have, ${tan}^{- 1} \frac{x - 1}{x - 2} + {tan}^{- 1} \frac{x + 1}{x + 2} = \frac{π}{4}$

$\Rightarrow {tan}^{- 1} \frac{x - 1}{x - 2} + {tan}^{- 1} \frac{x + 1}{x + 2} = {tan}^{- 1} 1$

$\Rightarrow {tan}^{- 1} \frac{x - 1}{x - 2} = {tan}^{- 1} 1 - {tan}^{- 1} \frac{x + 1}{x + 2}$

$\Rightarrow {tan}^{- 1} \frac{x - 1}{x - 2} = {tan}^{- 1} \frac{1 - \frac{x + 1}{x + 2}}{1 + \frac{x + 1}{x + 2}} = {tan}^{- 1} \frac{1}{2 x + 3}$

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

$\Rightarrow (x - 1) (2 x + 3) = x - 2$

$\Rightarrow 2 x^{2} + x - 3 = x - 2$

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

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Similar questions

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. If

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

then x is

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. Match the entries of Column - I and Column - II.
Number of real solutions of

	Column - I		Column - II
a	tan $(\frac{π}{4} + \frac{1}{2} c o s^{- 1} x)$ + tan $(\frac{π}{4} - \frac{1}{2} c o s^{- 1} x)$ = 1	p	0
b	$t a n^{- 1}$ $\frac{1}{2 x + 1}$ + $t a n^{- 1}$ $\frac{1}{4 x + 1}$ = $t a n^{- 1}$ $\frac{2}{x^{2}}$	q	2
c	$t a n^{- 1}$ $(x + \frac{2}{x})$ - $t a n^{- 1}$ $(x - \frac{2}{x})$ - $t a n^{- 1}$ $\frac{4}{x}$ = 0	r	3
d	$t a n^{- 1}$ (1 - x) + $t a n^{- 1}$ (1 + x) = $t a n^{- 1}$ 2x	s	1

Q. Solve :

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

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If tan−1x−1x−2+tan−1x+1x+2=π4, then x=

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