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Inverse Trigonometric Functions — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsInverse Trigonometric Functions3 Marks
Question
Write the value of $\tan^{-1}\frac{\text{a}}{\text{b}}-\tan^{-1}\Big(\frac{\text{a}-\text{b}}{\text{a}+\text{b}}\Big).$
✓

Answer

$\tan^{-1}\frac{\text{a}}{\text{b}}-\tan^{-1}\Big(\frac{\text{a}-\text{b}}{\text{a}+\text{b}}\Big)$
$=\tan^{-1}\begin{bmatrix}\frac{\frac{\text{a}}{\text{b}}-\frac{\text{a}-\text{b}}{\text{a}+\text{b}}}{1+\Big(\frac{\text{a}}{\text{b}}\Big)-\Big(\frac{\text{a}-\text{b}}{\text{a}+\text{b}}\Big)}\end{bmatrix}$
$\Big\{\text{Since},\tan^{-1}\text{x}-\tan^{-1}\text{y}=\tan^{-1}\Big(\frac{\text{x}-\text{y}}{1+\text{xy}}\Big)\Big\}$
$=\tan^{-1}\begin{bmatrix}\frac{\frac{\text{a}^2+\text{ab}-\text{ab}+\text{b}^2}{\text{b}(\text{a}+\text{b})}}{\frac{\text{ba}+\text{b}^2+\text{a}^2-\text{ab}}{{\text{b}(\text{a}+\text{b})}}}\end{bmatrix}$
$=\tan^{-1}\Big[\frac{\text{a}^2+\text{b}^2}{\text{a}^2+\text{b}^2}\Big]$
$=\tan^{-1}(1)$
$=\frac{\pi}{4}$
Hence,
$\tan^{-1}\frac{\text{a}}{\text{b}}-\tan^{-1}\Big(\frac{\text{a}-\text{b}}{\text{a}+\text{b}}\Big)=\frac{\pi}{4}$

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