Q.16 Prove that : tan⁻¹(1/2) + tan⁻¹(1/3) = π/4

Answer: Using tan⁻¹x + tan⁻¹y = tan⁻¹x + y/1 - xy (for xy lt 1 ), the L.H.S. equals tan⁻¹1 = π/4 . Here x = 1/2 , y = 1/3 , and xy = 1/6 lt 1 . L.H.S. = tan⁻¹(1/2 + 1/3/1 - 1/2·1/3) = tan⁻¹(5/6/5/6) = tan⁻¹(1) = π/4 = R.H.S.

Prove that : \(\tan^{-1}\left(\frac{1}{2}\right) + \tan^{-1}\left(\frac{1}{3}\right) = \frac{\pi}{4}\)

Q.16 Prove that : \(\tan^{-1}\left(\frac{1}{2}\right) + \tan^{-1}\left(\frac{1}{3}\right) = \frac{\pi}{4}\)

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

1 answer

Best answer

Answer: Using \(\tan^{-1}x + \tan^{-1}y = \tan^{-1}\frac{x + y}{1 - xy}\) (for \(xy \lt 1\)), the L.H.S. equals \(\tan^{-1}1 = \frac{\pi}{4}\).

Here \(x = \frac{1}{2}\), \(y = \frac{1}{3}\), and \(xy = \frac{1}{6} \lt 1\).

L.H.S. \(= \tan^{-1}\left(\frac{\frac{1}{2} + \frac{1}{3}}{1 - \frac{1}{2}\cdot\frac{1}{3}}\right) = \tan^{-1}\left(\frac{\frac{5}{6}}{\frac{5}{6}}\right) = \tan^{-1}(1) = \frac{\pi}{4}\) = R.H.S.

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

Was this answer helpful?

Prove that : \(\tan^{-1}\left(\frac{1}{2}\right) + \tan^{-1}\left(\frac{1}{3}\right) = \frac{\pi}{4}\)

1 answer

Your answer