Q.22 Find the approximate value of tan⁻¹(1.002) . [Given : π = 3.1416 ]

Answer: tan⁻¹(1.002) ≈ 0.7864 Let f(x) = tan⁻¹x . Then f'(x) = 1/1 + x² . Take a = 1 , h = 0.002 . f(a) = tan⁻¹1 = π/4 = 3.1416/4 = 0.7854 , and f'(a) = 1/1 + 1 = 1/2 . f(a + h) ≈ f(a) + h f'(a) = 0.7854 + 0.002 × 1/2 = 0.7854 + 0.001 = 0.7864 .

Find the approximate value of \(\tan^{-1}(1.002)\).[Given : \(\pi = 3.1416\)]

Q.22 Find the approximate value of \(\tan^{-1}(1.002)\).
[Given : \(\pi = 3.1416\)]

Doubtly Asked 29 Oct 2026 Board exam · March 2024 hsc 12th Save WhatsApp Report

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Answer: \(\tan^{-1}(1.002) \approx 0.7864\)

Let \(f(x) = \tan^{-1}x\). Then \(f'(x) = \frac{1}{1 + x^2}\).

Take \(a = 1\), \(h = 0.002\). \(f(a) = \tan^{-1}1 = \frac{\pi}{4} = \frac{3.1416}{4} = 0.7854\), and \(f'(a) = \frac{1}{1 + 1} = \frac{1}{2}\).

\(f(a + h) \approx f(a) + h f'(a) = 0.7854 + 0.002 \times \frac{1}{2} = 0.7854 + 0.001 = 0.7864\).

Doubtly AI · 29 Oct 2026 AI-generated · not yet verified Report

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Find the approximate value of \(\tan^{-1}(1.002)\).[Given : \(\pi = 3.1416\)]

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