The solution of tan -1 1-x -tan -1 1-x - 2 is-

The correct option is D x=0 Given, #160;tan ^ 1 ( 1+x ) #160; +tan ^ 1 ( 1 x ) #160; =π #160;/ 2 #160;⇒tan ^ 1 ( 1+x+1 x / 1 ( 1+x ) ...

266

You visited us 266 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Mathematics

Standard Formulae - 1

The solution ...

Question

The solution of ${tan}^{- 1} (1 + x) + {tan}^{- 1} (1 - x) = \frac{π}{2}$ is-

x = 1

No worries! We‘ve got your back. Try BYJU‘S free classes today!

x = - 1

No worries! We‘ve got your back. Try BYJU‘S free classes today!

x = 0

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

x = π

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is D $x = 0$
Given, ${tan}^{- 1} (1 + x) + {tan}^{- 1} (1 - x) = \frac{π}{2}$
$\Rightarrow {tan}^{- 1} (\frac{1 + x + 1 - x}{1 - (1 + x) (1 - x)}) = \frac{π}{2}$
$\Rightarrow \frac{2}{1 - (1 - x^{2})} = tan \frac{π}{2}$
$\Rightarrow \frac{2}{1 - 1 + x^{2}} = tan \frac{π}{2} \Rightarrow \frac{2}{x^{2}} = \frac{1}{0}$
$\Rightarrow x^{2} = 0 => x = 0$

Suggest Corrections

Similar questions

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. If

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

, then

x =

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