For the equation $2 x = tan ({tan}^{- 1} a) + 2 tan ({tan}^{- 1} a + {tan}^{- 1} a^{3})$ , which of the following is invalid?

a^{2} + 2 a = x

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a^{2} + 2 a x + 1 = 0

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a \neq 0

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a \neq - 1, 1

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Solution

The correct options are
A $a^{2} + 2 a = x$
B $a^{2} + 2 a x + 1 = 0$
C $a \neq 0$
$2 x = tan (2 {tan}^{- 1} a) + 2 tan ({tan}^{- 1} a + {tan}^{- 1} a^{3})$
$\Rightarrow 2 x = tan ({tan}^{- 1} (\frac{2 a}{1 - a^{2}})) + 2 tan ({tan}^{- 1} (\frac{a + a^{3}}{1 - a \cdot a^{3}}))$
$\Rightarrow 2 x = \frac{2 a}{1 - a^{2}} + 2 (\frac{a (1 + a^{2})}{1 - a^{4}})$
$\Rightarrow 2 x = \frac{2 a}{1 - a^{2}} + 2 (\frac{a (1 + a^{2})}{(1 - a^{2}) (1 + a^{2})})$
$\Rightarrow 2 x = a + \frac{2 a}{1 - a^{2}}$
$\Rightarrow x = \frac{2 a}{1 - a^{2}}; a^{2} \neq 1 \Rightarrow a \neq - 1, 1$
$\Rightarrow x (1 - a^{2}) = 2 a$
$\Rightarrow a^{2} x + 2 a - x = 0$
$\Rightarrow a^{2} x + 2 a = x$
Hence all are invalid except $(D)$ .

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