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বিপরীত ত্রিকোণমিতিক ফাংশনের ডোমেন ও রেঞ্জের ছক
ফাংশন ডোমেন রেঞ্জ \(\sin^{-1}x\) \([-1, 1]\) \(\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\) \(\cos^{-1}x\) \([-1, 1]\) \([0, \pi]\) \(\tan^{-1}x\) \((-\infty, \infty)\) বা \(\mathbb{R}\) \(\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)\) \(\cot^{-1}x\) \((-\infty, \infty)\) বা \(\mathbb{R}\) \((0, \pi)\) \(\sec^{-1}x\) \((-\infty, -1] \cup [1, +\infty)\) বা, \(\mathbb{R} - (-1, 1)\) \(\left[0, \frac{\pi}{2}\right) \cup \left[\pi, \frac{3\pi}{2}\right)\) \(\text{cosec}^{-1}x\) \((-\infty, -1] \cup [1, +\infty)\) বা, \(\mathbb{R} - (-1, 1)\) \(\left(0, \frac{\pi}{2}\right] \cup \left(\pi, \frac{3\pi}{2}\right]\) - \(\sin^{-1}x = \text{cosec}^{-1}\frac{1}{x} = \cos^{-1}\sqrt{1-x^2} = \sec^{-1}\frac{1}{\sqrt{1-x^2}} = \cot^{-1}\frac{\sqrt{1-x^2}}{x} = \tan^{-1}\frac{x}{\sqrt{1-x^2}}\)
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- \(\sin^{-1}x + \cos^{-1}x = \frac{\pi}{2}\)
- \(\tan^{-1}x + \cot^{-1}x = \frac{\pi}{2}\)
- \(\text{cosec}^{-1}x + \sec^{-1}x = \frac{\pi}{2}\)
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- \(\tan^{-1}x + \tan^{-1}y = \tan^{-1}\frac{x+y}{1-xy}\)
- \(\tan^{-1}x - \tan^{-1}y = \tan^{-1}\frac{x-y}{1+xy}\)
- \(\tan^{-1}x + \tan^{-1}y + \tan^{-1}z = \tan^{-1}\frac{x+y+z-xyz}{1-yz-zx-xy}\)
- \(\sin^{-1}x + \sin^{-1}y = \sin^{-1}\left\{x\sqrt{1-y^2} + y\sqrt{1-x^2}\right\}\); যখন \(x^2 + y^2 \le 1\)
- \(\sin^{-1}x - \sin^{-1}y = \sin^{-1}\left\{x\sqrt{1-y^2} - y\sqrt{1-x^2}\right\}\)
- \(\cos^{-1}x + \cos^{-1}y = \cos^{-1}\left\{xy - \sqrt{(1-x^2)(1-y^2)}\right\}\); যখন \(x + y \ge 0\)
- \(\cos^{-1}x - \cos^{-1}y = \cos^{-1}\left\{xy + \sqrt{(1-x^2)(1-y^2)}\right\}\)
- \(2\tan^{-1}x = \tan^{-1}\frac{2x}{1-x^2} = \sin^{-1}\frac{2x}{1+x^2} = \cos^{-1}\frac{1-x^2}{1+x^2}\)
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