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Inverse Trigonometric Functions — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsInverse Trigonometric Functions5 Marks
Question
For any a, b, x, y > 0, prove that:
$\frac{2}{3}\tan^{-1}\Big(\frac{3\text{a}\text{b}^2-\text{a}^3}{\text{b}^3-3\text{a}^2\text{b}}\Big)+\frac{2}{3}\tan^{-1}\Big(\frac{3\text{x}\text{y}^2-\text{x}^3}{\text{y}^3-3\text{x}^2\text{y}}\Big)=\tan^{-1}\frac{2\alpha\beta}{\alpha^2-\beta^2}$
where $\alpha=-\text{ax}+\text{by},\beta=\text{bx}+\text{ay}$
✓

Answer

Let $\text{a}=\text{b}\tan\text{m}$ and $\text{x}=\text{y}\tan\text{n}$
Then
$\frac{2}{3}\tan^{-1}\Big(\frac{3\text{a}\text{b}^2-\text{a}^3}{\text{b}^3-3\text{a}^2\text{b}}\Big)+\frac{2}{3}\tan^{-1}\Big(\frac{3\text{x}\text{y}^2-\text{x}^3}{\text{y}^3-3\text{x}^2\text{y}}\Big)$
$=\frac{2}{3}\tan^{-1}\Big(\frac{3\text{b}^3\tan\text{m}-\text{b}^3\tan^3\text{m}}{\text{b}^3-3\text{b}^3\tan^2\text{m}}\Big)+\frac{2}{3}\tan^{-1}\Big(\frac{3\text{y}^3\tan\text{n}-\text{y}^3\tan^3\text{n}}{\text{y}^3-3\text{y}^3\tan^2\text{n}}\Big)$
$=\frac{2}{3}\tan^{-1}\Big(\frac{3\tan\text{m}-\tan^3\text{m}}{1-3\tan^2\text{m}}\Big)+\frac{2}{3}\tan^{-1}\Big(\frac{3\tan\text{n}-\tan^3\text{n}}{1-\tan^2\text{n}}\Big)$
$=\frac{2}{3}\tan^{-1}(\tan3\text{m})+\frac{2}{3}\tan^{-1}(\tan3\text{n})$ $[\because\ \text{a}=\text{b}\tan\text{m},\text{x}=\text{y}=\tan\text{n}]$
$=2\tan^{-1}\Bigg(\frac{\frac{\text{a}}{\text{b}}+\frac{\text{x}}{\text{y}}}{1-\frac{\text{a}}{\text{b}}\frac{\text{x}}{\text{y}}}\Bigg)$
$=2\tan^{-1}\Big(\frac{\text{ay}+\text{bx}}{\text{by}-\text{ax}}\Big)$
$=\tan^{-1}\begin{Bmatrix}\frac{2\frac{\text{ay}+\text{bx}}{\text{by}-\text{ax}}}{1-\Big(\frac{\text{ay}+\text{bx}}{\text{by}-\text{ax}}\Big)^2}\end{Bmatrix}$
$=\tan^{-1}\Bigg\{\frac{2(\text{ay}+\text{bx})(\text{by}-\text{ax})}{(\text{by}-\text{ax})^2-(\text{ay}+\text{bx})^2}\Bigg\}$
$=\tan^{-1}\Big\{\frac{2\alpha\beta}{\alpha^2-\beta^2}\Big\}$ $[\because\ \beta=\text{bx}+\text{ay},\alpha=-\text{ax}+\text{by}]$

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