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Inverse Trigonometric Functions — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsInverse Trigonometric Functions5 Marks
Question
If $\sin^{-1}\frac{2\text{a}}{1+\text{a}^2}-\cos^{-1}\frac{1-\text{b}^2}{1+\text{b}^2}=\tan^{-1}\frac{2\text{x}}{1-\text{x}^2},$ then prove that $\text{x}=\frac{\text{a}-\text{b}}{1+\text{ab}}$
✓

Answer

Let:
$\text{a}=\tan\text{m}$
$\text{b}=\tan\text{n}$
$\text{x}=\tan\text{y}$
Now,
$\sin^{-1}\frac{2\text{a}}{1+\text{a}^2}-\cos^{-1}\frac{1-\text{b}^2}{1+\text{b}^2}=\tan^{-1}\frac{2\text{x}}{1-\text{x}^2},$
$\Rightarrow\sin^{-1}\frac{2\tan\text{m}}{1+\tan^2\text{m}}-\cos^{-1}\frac{1-\tan^{2}\text{n}}{1+\tan^{2}\text{n}}=\tan^{-1}\frac{2\tan\text{y}}{1-\tan^2\text{y}}$
$\Rightarrow\sin^{-1}(\sin2\text{m})-\cos^{-1}(\cos2\text{n})=\tan^{-1}(\tan2\text{y})$
$\Big[\because\ \sin2\text{x}=\frac{2\tan\text{x}}{1+\tan^2\text{x}},\cos2\text{x}=\frac{1-\tan^2\text{x}}{1+\tan^2\text{x}}\Big]$
$\Rightarrow2\text{m}-2\text{n}=2\text{y}$
$\Rightarrow\text{m}-\text{n}=\text{y}$
$\Rightarrow\tan^{-1}\text{a}-\tan^{-1}\text{b}=\tan^{-1}\text{x}$
$[\because\ \text{a}=\tan\text{m},b=\tan\text{n}\text{ and }\text{x}=\tan\text{y}]$
$\Rightarrow\tan^{-1}\frac{\text{a}-\text{b}}{1+\text{ab}}=\tan^{-1}\text{x}$
$\Big[\because\ \tan^{-1}\text{x}-\tan^{-1}\text{y}=\tan^{-1}\frac{\text{x}-\text{y}}{1+\text{xy}}\Big]$
$\Rightarrow\frac{\text{a}-\text{b}}{1+\text{ab}}=\text{x}$

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