Question

If $\tan ^ { - 1 } 2 x + \tan ^ { - 1 } 3 x = \frac { \pi } { 4 }$, then x =

• A. – 1
• B. $\frac { 1 } { 6 }$
• C. $- 1 , \frac { 1 } { 6 }$
• D. None of these

Answer 1

Answer: (B)

$\frac { 1 } { 6 }$

Answer 2

$\tan ^ { - 1 } 2 x + \tan ^ { - 1 } 3 x = \frac { \pi } { 4 }$ $\Rightarrow \tan ^ { - 1 } \left( \frac { 2 x + 3 x } { 1 - ( 2 x ) ( 3 x ) } \right) = \frac { \pi } { 4 }$

$\Rightarrow \tan ^ { - 1 } \left( \frac { 5 x } { 1 - 6 x ^ { 2 } } \right) = \tan ^ { - 1 } ( 1 )$ $\Rightarrow \frac { 5 x } { 1 - 6 x ^ { 2 } } = 1$

$\Rightarrow 1 - 6 x ^ { 2 } = 5 x$ $\Rightarrow 6 x ^ { 2 } + 5 x - 1 = 0$

$\Rightarrow ( x + 1 ) \left( x - \frac { 1 } { 6 } \right) = 0$ $\Rightarrow x = - 1 , \frac { 1 } { 6 }$

But – 1 does not hold.

Trick : Check with the options. Obviously the equation holds for $x = \frac { 1 } { 6 }$ , but not for – 1.