Ravi Kant Kumar

Prove 2sin⁻¹(3/5) − tan⁻¹(17/31) = π/4 Prove that \( 2\sin^{-1}\left(\frac{3}{5}\right) – \tan^{-1}\left(\frac{17}{31}\right) = \frac{\pi}{4} \) Solution: Let \[ \theta = \sin^{-1}\left(\frac{3}{5}\right) \Rightarrow \sin \theta = \frac{3}{5} \] Then, \[ \cos \theta = \frac{4}{5} \Rightarrow \tan \theta = \frac{3}{4} \] Using identity: \[ \tan(2\theta) = \frac{2\tan\theta}{1 – \tan^2\theta} \] \[ = \frac{2 \cdot \frac{3}{4}}{1 – \left(\frac{3}{4}\right)^2} […]

Prove sin⁻¹(4/5) + 2tan⁻¹(1/3) = π/2 Prove that \( \sin^{-1}\left(\frac{4}{5}\right) + 2\tan^{-1}\left(\frac{1}{3}\right) = \frac{\pi}{2} \) Solution: Let \[ \theta = \tan^{-1}\left(\frac{1}{3}\right) \] Then, \[ \tan \theta = \frac{1}{3} \] Using double angle identity: \[ \tan(2\theta) = \frac{2\tan\theta}{1 – \tan^2\theta} \] \[ = \frac{2 \cdot \frac{1}{3}}{1 – \frac{1}{9}} = \frac{2/3}{8/9} = \frac{3}{4} \] \[ \Rightarrow 2\theta

Prove tan⁻¹(1/7) + 2tan⁻¹(1/3) = π/4 Prove that \( \tan^{-1}\left(\frac{1}{7}\right) + 2\tan^{-1}\left(\frac{1}{3}\right) = \frac{\pi}{4} \) Solution: Let \[ \theta = \tan^{-1}\left(\frac{1}{3}\right) \] Then, \[ \tan \theta = \frac{1}{3} \] Using double angle identity: \[ \tan(2\theta) = \frac{2\tan\theta}{1 – \tan^2\theta} \] \[ = \frac{2 \cdot \frac{1}{3}}{1 – \left(\frac{1}{3}\right)^2} \] \[ = \frac{2/3}{1 – 1/9} = \frac{2/3}{8/9}

Prove tan⁻¹(2/3) = ½tan⁻¹(12/5) Prove that \( \tan^{-1}\left(\frac{2}{3}\right) = \frac{1}{2}\tan^{-1}\left(\frac{12}{5}\right) \) Solution: Let \[ \theta = \tan^{-1}\left(\frac{2}{3}\right) \] Then, \[ \tan \theta = \frac{2}{3} \] Using identity: \[ \tan(2\theta) = \frac{2\tan\theta}{1 – \tan^2\theta} \] \[ = \frac{2 \cdot \frac{2}{3}}{1 – \left(\frac{2}{3}\right)^2} \] \[ = \frac{4/3}{1 – 4/9} \] \[ = \frac{4/3}{5/9} \] \[ = \frac{4}{3}

Prove tan⁻¹(1/4) + tan⁻¹(2/9) = ½cos⁻¹(3/5) Prove that \( \tan^{-1}\left(\frac{1}{4}\right) + \tan^{-1}\left(\frac{2}{9}\right) = \frac{1}{2}\cos^{-1}\left(\frac{3}{5}\right) = \frac{1}{2}\sin^{-1}\left(\frac{4}{5}\right) \) Solution: Let \[ \theta = \tan^{-1}\left(\frac{1}{4}\right) + \tan^{-1}\left(\frac{2}{9}\right) \] Using identity: \[ \tan(A+B) = \frac{\tan A + \tan B}{1 – \tan A \tan B} \] \[ \tan \theta = \frac{\frac{1}{4} + \frac{2}{9}}{1 – \frac{1}{4}\cdot\frac{2}{9}} \] \[ = \frac{\frac{9

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