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INVERSE TRIGNOMETRIC FUNCTIONS — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsINVERSE TRIGNOMETRIC FUNCTIONS3 Marks
Question
Prove the following:$\tan^{-1}\text{x}+\tan^{-1}\Bigg(\frac{\text{2x}}{\text{1 - x}^{2}}\Bigg)=\tan^{-1}\Bigg(\frac{\text{3x - x}^{2}}{\text{1 - 3x}^{2}}\Bigg)$.
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Answer

$LHS = \tan^{-1} \Bigg[\frac{\text{x}+\frac{\text{2x}}{\text{1 - x}^{2}}}{{\text{1 -x}\frac{\text{2x}}{\text{1 - x}^{2}}}}\Bigg]$
$= \tan^{-1} \Bigg[\frac{\text{x(1 - x}^{2})+\text{2x}}{\text{1 - x}^{2}-\text{2x}^{2}}\Bigg]$
$= \tan^{-1} \Bigg[\frac{\text{3x - x}^{3}}{\text{1 - 3x}^{2}}\Bigg]=\text{RHS}.$

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