Write the following in the simplest form: ^ -1 a x^2-a^2 ,|x|>a - Vidyadip

Question

Write the following in the simplest form:
$\cot^{-1}\frac{\text{a}}{\sqrt{\text{x}^2-\text{a}^2}},|\text{x}|>\text{a}$

✓

Answer

$\cot^{-1}\frac{\text{a}}{\sqrt{\text{x}^2-\text{a}^2}},|\text{x}|>\text{a}$ Let, $\text{x}=\text{a}\sec\theta$ $\cot^{-1}\bigg(\frac{\text{a}}{\sqrt{\text{a}^2\sec^{2}\theta-\text{a}^2}}\bigg)$ $=\cot^{-1}\begin{pmatrix}\frac{\text{a}}{\sqrt{\text{a}^2\big(\sec^{2}\theta-1\big)}}\end{pmatrix}$ $=\cot^{-1}\frac{1}{\sqrt{\tan^2\theta}}$ $\{\text{since},\sec^2\theta-1=\tan^2\theta\}$ $=\cot^{-1}(\cot\theta)$ $=\theta$ $=\sec^{-1}\Big(\frac{\text{x}}{\text{a}}\Big)$Hence,
$\cot^{-1}\frac{a}{\sqrt{\text{x}^2-\text{a}^2}}=\sec^{-1}\Big(\frac{\text{x}}{\text{a}}\Big)$

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