Ravi Kant Kumar

Prove cos⁻¹(4/5) + cos⁻¹(12/13) = cos⁻¹(33/65) Problem Prove: \( \cos^{-1}\left(\frac{4}{5}\right) + \cos^{-1}\left(\frac{12}{13}\right) = \cos^{-1}\left(\frac{33}{65}\right) \) Solution Let: \[ A = \cos^{-1}\left(\frac{4}{5}\right), \quad B = \cos^{-1}\left(\frac{12}{13}\right) \] Step 1: Find sin A and sin B \[ \cos A = \frac{4}{5} \Rightarrow \sin A = \frac{3}{5} \] \[ \cos B = \frac{12}{13} \Rightarrow \sin B = \frac{5}{13} […]

Solve cos⁻¹(√3x) + cos⁻¹(x) = π/2 Problem Solve: \( \cos^{-1}(\sqrt{3}x) + \cos^{-1}(x) = \frac{\pi}{2} \) Solution Step 1: Use identity If: \[ \cos^{-1}(a) + \cos^{-1}(b) = \frac{\pi}{2} \] then: \[ a^2 + b^2 = 1 \] Step 2: Apply \[ (\sqrt{3}x)^2 + x^2 = 1 \] \[ 3x^2 + x^2 = 1 \] \[ 4x^2

Prove 9x² − 12xy cosα + 4y² = 36 sin²α Problem If \[ \cos^{-1}\left(\frac{x}{2}\right) + \cos^{-1}\left(\frac{y}{3}\right) = \alpha \] prove that \[ 9x^2 – 12xy\cos\alpha + 4y^2 = 36\sin^2\alpha \] Solution Let: \[ A = \cos^{-1}\left(\frac{x}{2}\right), \quad B = \cos^{-1}\left(\frac{y}{3}\right) \] \[ A + B = \alpha \] Step 1: Express cos A, cos B

Prove 9x² − 12xy cosα + 4y² = 36 sin²α Problem If \[ \cos^{-1}\left(\frac{x}{2}\right) + \cos^{-1}\left(\frac{y}{3}\right) = \alpha \] prove that \[ 9x^2 – 12xy\cos\alpha + 4y^2 = 36\sin^2\alpha \] Solution Let: \[ A = \cos^{-1}\left(\frac{x}{2}\right), \quad B = \cos^{-1}\left(\frac{y}{3}\right) \] \[ A + B = \alpha \] Step 1: Express cos A, cos B

Solve cos⁻¹(x) + sin⁻¹(x/2) − π/6 = 0 Problem Solve: \( \cos^{-1}(x) + \sin^{-1}\left(\frac{x}{2}\right) – \frac{\pi}{6} = 0 \) Solution Step 1: Rearrange \[ \cos^{-1}(x) + \sin^{-1}\left(\frac{x}{2}\right) = \frac{\pi}{6} \] Step 2: Convert cos⁻¹ into sin⁻¹ \[ \cos^{-1}(x) = \frac{\pi}{2} – \sin^{-1}(x) \] Step 3: Substitute \[ \frac{\pi}{2} – \sin^{-1}(x) + \sin^{-1}\left(\frac{x}{2}\right) = \frac{\pi}{6} \]

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) –