April 2026

Solve sin⁻¹x + sin⁻¹(2x) = π/3 Problem Solve: \( \sin^{-1}(x) + \sin^{-1}(2x) = \frac{\pi}{3} \) Solution Let: \[ A = \sin^{-1}(x), \quad B = \sin^{-1}(2x) \] Step 1: Use identity \[ \sin(A+B) = \sin A \cos B + \cos A \sin B \] \[ \sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2} \] Step 2: Express terms \[ \sin A […]

Prove (9π/8) − (9/4)sin⁻¹(1/3) = (9/4)sin⁻¹(2√2/3) Problem Prove: \[ \frac{9\pi}{8} – \frac{9}{4}\sin^{-1}\left(\frac{1}{3}\right) = \frac{9}{4}\sin^{-1}\left(\frac{2\sqrt{2}}{3}\right) \] Solution Step 1: Factor common term \[ = \frac{9}{4}\left(\frac{\pi}{2} – \sin^{-1}\left(\frac{1}{3}\right)\right) \] Step 2: Use identity \[ \frac{\pi}{2} – \sin^{-1}x = \cos^{-1}x \] \[ = \frac{9}{4}\cos^{-1}\left(\frac{1}{3}\right) \] Step 3: Show equivalence Let: \[ \theta = \cos^{-1}\left(\frac{1}{3}\right) \] \[ \cos\theta =

Prove sin⁻¹(5/13) + cos⁻¹(3/5) = tan⁻¹(63/16) Problem Prove: \( \sin^{-1}\left(\frac{5}{13}\right) + \cos^{-1}\left(\frac{3}{5}\right) = \tan^{-1}\left(\frac{63}{16}\right) \) Solution Let: \[ A = \sin^{-1}\left(\frac{5}{13}\right), \quad B = \cos^{-1}\left(\frac{3}{5}\right) \] Step 1: Find tan A \[ \sin A = \frac{5}{13} \Rightarrow \cos A = \frac{12}{13} \Rightarrow \tan A = \frac{5}{12} \] Step 2: Find tan B \[ \cos B

Prove sin⁻¹(63/65) = sin⁻¹(5/13) + cos⁻¹(3/5) Problem Prove: \( \sin^{-1}\left(\frac{63}{65}\right) = \sin^{-1}\left(\frac{5}{13}\right) + \cos^{-1}\left(\frac{3}{5}\right) \) Solution Let: \[ A = \sin^{-1}\left(\frac{5}{13}\right), \quad B = \cos^{-1}\left(\frac{3}{5}\right) \] Step 1: Find sin A, cos A \[ \sin A = \frac{5}{13} \Rightarrow \cos A = \frac{12}{13} \] Step 2: Find sin B, cos B \[ \cos B =

Sum of Series tan⁻¹ Terms Problem Sum the series: \[ \tan^{-1}\left(\frac{1}{3}\right) + \tan^{-1}\left(\frac{4}{33}\right) + \cdots + \tan^{-1}\left(\frac{2^{\,n-1}}{1+2^{\,2n-1}}\right) \] Solution Step 1: Observe pattern Each term can be written using identity: \[ \tan^{-1}a – \tan^{-1}b = \tan^{-1}\left(\frac{a-b}{1+ab}\right) \] Step 2: Apply identity Let: \[ a = 2^k,\quad b = 2^{k-1} \] \[ \tan^{-1}\left(\frac{2^{k-1}}{1+2^{2k-1}}\right) = \tan^{-1}(2^k) –

Evaluate cos(sin⁻¹(3/5) + sin⁻¹(5/13)) Problem Evaluate: \( \cos\left(\sin^{-1}\left(\frac{3}{5}\right) + \sin^{-1}\left(\frac{5}{13}\right)\right) \) Solution Let: \[ A = \sin^{-1}\left(\frac{3}{5}\right), \quad B = \sin^{-1}\left(\frac{5}{13}\right) \] Step 1: Find cos A and cos B \[ \sin A = \frac{3}{5} \Rightarrow \cos A = \frac{4}{5} \] \[ \sin B = \frac{5}{13} \Rightarrow \cos B = \frac{12}{13} \] Step 2: Use

Solve tan⁻¹((x−2)/(x−1)) + tan⁻¹((x+2)/(x+1)) = π/4 Problem Solve: \( \tan^{-1}\left(\frac{x-2}{x-1}\right) + \tan^{-1}\left(\frac{x+2}{x+1}\right) = \frac{\pi}{4} \) Solution Let: \[ A = \tan^{-1}\left(\frac{x-2}{x-1}\right), \quad B = \tan^{-1}\left(\frac{x+2}{x+1}\right) \] 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-2}{x-1} +

Solve tan⁻¹(2+x) + tan⁻¹(2−x) = tan⁻¹(2/3) Problem Solve: \( \tan^{-1}(2+x) + \tan^{-1}(2-x) = \tan^{-1}\left(\frac{2}{3}\right) \), where \( x < -\sqrt{3} \) or \( x > \sqrt{3} \) Solution Let: \[ A = \tan^{-1}(2+x), \quad B = \tan^{-1}(2-x) \] Step 1: Use tan(A + B) \[ \tan(A + B) = \frac{\tan A + \tan B}{1 –

Solve tan⁻¹((x−2)/(x−4)) + tan⁻¹((x+2)/(x+4)) = π/4 Problem Solve: \( \tan^{-1}\left(\frac{x-2}{x-4}\right) + \tan^{-1}\left(\frac{x+2}{x+4}\right) = \frac{\pi}{4} \) Solution Let: \[ A = \tan^{-1}\left(\frac{x-2}{x-4}\right), \quad B = \tan^{-1}\left(\frac{x+2}{x+4}\right) \] 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-2}{x-4} +

Solve tan⁻¹(x/2) + tan⁻¹(x/3) = π/4 Problem Solve: \( \tan^{-1}\left(\frac{x}{2}\right) + \tan^{-1}\left(\frac{x}{3}\right) = \frac{\pi}{4}, \quad 0 < x < \sqrt{6} \) Solution Let: \[ A = \tan^{-1}\left(\frac{x}{2}\right), \quad B = \tan^{-1}\left(\frac{x}{3}\right) \] Step 1: Use tan(A + B) \[ \tan(A + B) = \frac{\tan A + \tan B}{1 – \tan A \tan B} \] \[