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On simplifying $\cot \left(2 \tan ^{-1} x\right)$ where $|x|<1$, we get: |
$\frac{1-x^2}{2 x}$ $\frac{2 x}{1-x^2}$ $\frac{1+x^2}{2 x}$ $\frac{2 x}{1+x^2}$ |
$\frac{1-x^2}{2 x}$ |
The correct answer is Option (1) → $\frac{1-x^2}{2 x}$ $\cot \left(2 \tan ^{-1} x\right)=\frac{1}{\tan\left(2 \tan ^{-1} x\right)}$ $=\frac{1}{\tan\left(\tan ^{-1}(\frac{2x}{1-x^2})\right)}=\frac{1-x^2}{2 x}$ |
