Ravi Kant Kumar

Solve tan⁻¹(x−1) + tan⁻¹(x) + tan⁻¹(x+1) = tan⁻¹(3x) Problem Solve: \( \tan^{-1}(x-1) + \tan^{-1}(x) + \tan^{-1}(x+1) = \tan^{-1}(3x) \) Solution Step 1: Combine first two terms \[ \tan^{-1}(x-1) + \tan^{-1}(x) = \tan^{-1}\left(\frac{(x-1)+x}{1 – x(x-1)}\right) \] \[ = \tan^{-1}\left(\frac{2x-1}{1 – x^2 + x}\right) \] Step 2: Add third term \[ \tan^{-1}\left(\frac{2x-1}{1 – x^2 + x}\right) + […]

Solve tan⁻¹(x+1) + tan⁻¹(x−1) = tan⁻¹(8/31) Problem Solve: \( \tan^{-1}(x+1) + \tan^{-1}(x-1) = \tan^{-1}\left(\frac{8}{31}\right) \) Solution Let: \[ A = \tan^{-1}(x+1), \quad B = \tan^{-1}(x-1) \] Step 1: Use tan(A + B) \[ \tan(A + B) = \frac{\tan A + \tan B}{1 – \tan A \tan B} \] \[ = \frac{(x+1) + (x-1)}{1 – (x+1)(x-1)}

Solve tan⁻¹(2x) + tan⁻¹(3x) = nπ + 3π/4 Problem Solve: \( \tan^{-1}(2x) + \tan^{-1}(3x) = n\pi + \frac{3\pi}{4} \) Solution Let: \[ A = \tan^{-1}(2x), \quad B = \tan^{-1}(3x) \] Step 1: Use tan(A + B) \[ \tan(A + B) = \frac{\tan A + \tan B}{1 – \tan A \tan B} \] \[ = \frac{2x

Evaluate tan⁻¹(x/y) − tan⁻¹((x−y)/(x+y)) Problem Find the value of \( \tan^{-1}\left(\frac{x}{y}\right) – \tan^{-1}\left(\frac{x-y}{x+y}\right) \) Solution Let: \[ A = \tan^{-1}\left(\frac{x}{y}\right), \quad B = \tan^{-1}\left(\frac{x-y}{x+y}\right) \] Step 1: Use tan(A − B) \[ \tan(A – B) = \frac{\tan A – \tan B}{1 + \tan A \tan B} \] Step 2: Substitute \[ = \frac{\frac{x}{y} – \frac{x-y}{x+y}}{1

Prove tan⁻¹(1/4) + tan⁻¹(2/9) = sin⁻¹(1/√5) Problem Prove: \( \tan^{-1}\left(\frac{1}{4}\right) + \tan^{-1}\left(\frac{2}{9}\right) = \sin^{-1}\left(\frac{1}{\sqrt{5}}\right) \) Solution Let: \[ A = \tan^{-1}\left(\frac{1}{4}\right), \quad B = \tan^{-1}\left(\frac{2}{9}\right) \] Step 1: Use tan(A + B) \[ \tan(A + B) = \frac{\tan A + \tan B}{1 – \tan A \tan B} \] \[ = \frac{\frac{1}{4} + \frac{2}{9}}{1 – \frac{1}{4}\cdot\frac{2}{9}}

Prove tan⁻¹(1/7) + tan⁻¹(1/13) = tan⁻¹(2/9) Problem Prove: \( \tan^{-1}\left(\frac{1}{7}\right) + \tan^{-1}\left(\frac{1}{13}\right) = \tan^{-1}\left(\frac{2}{9}\right) \) Solution Let: \[ A = \tan^{-1}\left(\frac{1}{7}\right), \quad B = \tan^{-1}\left(\frac{1}{13}\right) \] Step 1: Use tan(A + B) identity \[ \tan(A + B) = \frac{\tan A + \tan B}{1 – \tan A \tan B} \] Step 2: Substitute values \[ =

Prove sin⁻¹(12/13) + cos⁻¹(4/5) + tan⁻¹(63/16) = π Problem Prove: \( \sin^{-1}\left(\frac{12}{13}\right) + \cos^{-1}\left(\frac{4}{5}\right) + \tan^{-1}\left(\frac{63}{16}\right) = \pi \) Solution Let: \[ A = \sin^{-1}\left(\frac{12}{13}\right), \quad B = \cos^{-1}\left(\frac{4}{5}\right) \] Step 1: Find tan A \[ \sin A = \frac{12}{13} = \frac{\text{Perpendicular}}{\text{Hypotenuse}} \] Perpendicular = 12 Hypotenuse = 13 Base: \[ \sqrt{13^2 – 12^2} =

Solve 5tan⁻¹x + 3cot⁻¹x = 2π Problem Solve: \( 5\tan^{-1}x + 3\cot^{-1}x = 2\pi \) Solution Step 1: Use identity \[ \cot^{-1}x = \frac{\pi}{2} – \tan^{-1}x \] Step 2: Substitute \[ 5\tan^{-1}x + 3\left(\frac{\pi}{2} – \tan^{-1}x\right) = 2\pi \] \[ 5\tan^{-1}x + \frac{3\pi}{2} – 3\tan^{-1}x = 2\pi \] \[ 2\tan^{-1}x + \frac{3\pi}{2} = 2\pi \]

Solve tan⁻¹x + 2cot⁻¹x = 2π/3 Problem Solve: \( \tan^{-1}x + 2\cot^{-1}x = \frac{2\pi}{3} \) Solution Step 1: Use identity \[ \cot^{-1}x = \frac{\pi}{2} – \tan^{-1}x \] Step 2: Substitute \[ \tan^{-1}x + 2\left(\frac{\pi}{2} – \tan^{-1}x\right) = \frac{2\pi}{3} \] \[ \tan^{-1}x + \pi – 2\tan^{-1}x = \frac{2\pi}{3} \] \[ \pi – \tan^{-1}x = \frac{2\pi}{3} \]

Solve 4sin⁻¹x = π − cos⁻¹x Problem Solve: \( 4\sin^{-1}x = \pi – \cos^{-1}x \) Solution Step 1: Use identity \[ \cos^{-1}x = \frac{\pi}{2} – \sin^{-1}x \] Step 2: Substitute \[ 4\sin^{-1}x = \pi – \left(\frac{\pi}{2} – \sin^{-1}x\right) \] \[ 4\sin^{-1}x = \frac{\pi}{2} + \sin^{-1}x \] Step 3: Solve \[ 3\sin^{-1}x = \frac{\pi}{2} \] \[