April 2026

Evaluate sin(2tan⁻¹(2/3)) + cos(tan⁻¹√3) Evaluate \( \sin\left(2\tan^{-1}\left(\frac{2}{3}\right)\right) + \cos\left(\tan^{-1}\sqrt{3}\right) \) Solution: Let \[ \theta = \tan^{-1}\left(\frac{2}{3}\right) \] Then, \[ \tan \theta = \frac{2}{3} \] Using identity: \[ \sin(2\theta) = \frac{2\tan\theta}{1 + \tan^2\theta} \] \[ = \frac{2 \cdot \frac{2}{3}}{1 + \left(\frac{2}{3}\right)^2} \] \[ = \frac{4/3}{1 + 4/9} \] \[ = \frac{4/3}{13/9} \] \[ = \frac{4}{3} \times […]

Evaluate sin(½ cos⁻¹(4/5)) Evaluate \( \sin\left(\frac{1}{2}\cos^{-1}\left(\frac{4}{5}\right)\right) \) Solution: Let \[ \theta = \cos^{-1}\left(\frac{4}{5}\right) \] Then, \[ \cos \theta = \frac{4}{5} \] Using identity: \[ \sin\left(\frac{\theta}{2}\right) = \sqrt{\frac{1 – \cos\theta}{2}} \] \[ = \sqrt{\frac{1 – \frac{4}{5}}{2}} \] \[ = \sqrt{\frac{\frac{1}{5}}{2}} \] \[ = \sqrt{\frac{1}{10}} \] \[ = \frac{1}{\sqrt{10}} \] Final Answer: \[ \sin\left(\frac{1}{2}\cos^{-1}\left(\frac{4}{5}\right)\right) = \frac{1}{\sqrt{10}} \]

Evaluate tan(½ sin⁻¹(3/4)) Evaluate \( \tan\left(\frac{1}{2}\sin^{-1}\left(\frac{3}{4}\right)\right) \) Solution: Let \[ \theta = \sin^{-1}\left(\frac{3}{4}\right) \] Then, \[ \sin \theta = \frac{3}{4} \] Consider a right triangle: Opposite = 3 Hypotenuse = 4 So, \[ \cos \theta = \sqrt{1 – \sin^2\theta} = \frac{\sqrt{7}}{4} \] Using half-angle identity: \[ \tan\left(\frac{\theta}{2}\right) = \frac{1 – \cos\theta}{\sin\theta} \] \[ = \frac{1

Evaluate tan{2tan⁻¹(1/5) − π/4} Evaluate \( \tan\left(2\tan^{-1}\left(\frac{1}{5}\right) – \frac{\pi}{4}\right) \) Solution: Let \[ \theta = \tan^{-1}\left(\frac{1}{5}\right) \] Then, \[ \tan \theta = \frac{1}{5} \] Using identity: \[ \tan(2\theta) = \frac{2\tan\theta}{1 – \tan^2\theta} \] \[ = \frac{2 \cdot \frac{1}{5}}{1 – \left(\frac{1}{5}\right)^2} \] \[ = \frac{2/5}{1 – 1/25} \] \[ = \frac{2/5}{24/25} \] \[ = \frac{2}{5} \times

Prove that 2sin⁻¹(3/5) = tan⁻¹(24/7) Prove that \(2\sin^{-1}\left(\frac{3}{5}\right) = \tan^{-1}\left(\frac{24}{7}\right)\) Solution: Let \[ \theta = \sin^{-1}\left(\frac{3}{5}\right) \] Then, \[ \sin \theta = \frac{3}{5} \] Consider a right triangle: Opposite = 3 Hypotenuse = 5 So, \[ \cos \theta = \frac{4}{5} \] Now, we use the identity: \[ \tan(2\theta) = \frac{2\tan\theta}{1 – \tan^2\theta} \] First find

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 \[ \frac{\pi}{2} – \sin^{-1}(x) + \sin^{-1}\left(\frac{x}{2}\right) = \frac{\pi}{6} \]