Ravi Kant Kumar

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) \[ \tan(A + B) = \frac{\tan A + \tan B}{1 – \tan A \tan B} \] \[

Solve tan⁻¹(x+2) + tan⁻¹(x−2) = tan⁻¹(8/79) 0″> Problem Solve: \( \tan^{-1}(x+2) + \tan^{-1}(x-2) = \tan^{-1}\left(\frac{8}{79}\right), \quad x > 0 \) Solution Let: \[ A = \tan^{-1}(x+2), \quad B = \tan^{-1}(x-2) \] Step 1: Use tan(A + B) \[ \tan(A + B) = \frac{\tan A + \tan B}{1 – \tan A \tan B} \] \[ =

Solve cot⁻¹x − cot⁻¹(x+2) = π/12 0″> Problem Solve: \( \cot^{-1}x – \cot^{-1}(x+2) = \frac{\pi}{12}, \quad x > 0 \) Solution Let: \[ A = \cot^{-1}x,\quad B = \cot^{-1}(x+2) \] Step 1: Use identity \[ \cot^{-1}x – \cot^{-1}y = \tan^{-1}\left(\frac{y – x}{1 + xy}\right) \] So, \[ \cot^{-1}x – \cot^{-1}(x+2) = \tan^{-1}\left(\frac{(x+2) – x}{1 +

Solve tan⁻¹((1−x)/(1+x)) − ½tan⁻¹(x) = 0 0″> Problem Solve: \( \tan^{-1}\left(\frac{1-x}{1+x}\right) – \frac{1}{2}\tan^{-1}(x) = 0,\quad x > 0 \) Solution Step 1: Rearrange \[ \tan^{-1}\left(\frac{1-x}{1+x}\right) = \frac{1}{2}\tan^{-1}(x) \] Step 2: Use identity \[ \tan^{-1}\left(\frac{1-x}{1+x}\right) = \frac{\pi}{4} – \tan^{-1}(x) \quad (x>0) \] Step 3: Substitute \[ \frac{\pi}{4} – \tan^{-1}(x) = \frac{1}{2}\tan^{-1}(x) \] Step 4: Solve \[

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)}