Ravi Kant Kumar

Find tan⁻¹{2cos(2sin⁻¹(1/2))} Evaluate \( \tan^{-1}\left\{2\cos\left(2\sin^{-1}\left(\frac{1}{2}\right)\right)\right\} \) Solution: Let \[ \theta = \sin^{-1}\left(\frac{1}{2}\right) \Rightarrow \theta = \frac{\pi}{6} \] Now, \[ 2\theta = \frac{\pi}{3} \] \[ \cos(2\theta) = \cos\left(\frac{\pi}{3}\right) = \frac{1}{2} \] Thus, \[ 2\cos(2\theta) = 2 \cdot \frac{1}{2} = 1 \] Hence, \[ \tan^{-1}(1) = \frac{\pi}{4} \] Final Answer: \[ \tan^{-1}\left\{2\cos\left(2\sin^{-1}\left(\frac{1}{2}\right)\right)\right\} = \frac{\pi}{4} \] Next Question […]

Show 2tan⁻¹x + sin⁻¹(2x/(1+x²)) is Constant Show that \( 2\tan^{-1}x + \sin^{-1}\left(\frac{2x}{1+x^2}\right) \) is constant for \( x \ge 1 \), and find its value. Solution: Let \[ \theta = \tan^{-1}(x) \Rightarrow x = \tan \theta \] Then, \[ \frac{2x}{1+x^2} = \frac{2\tan\theta}{1+\tan^2\theta} = \sin(2\theta) \] Thus, \[ \sin^{-1}\left(\frac{2x}{1+x^2}\right) = \sin^{-1}(\sin 2\theta) \] Since \( x

Prove x = (a+b)/(1−ab) If \( \sin^{-1}\left(\frac{2a}{1+a^2}\right) + \sin^{-1}\left(\frac{2b}{1+b^2}\right) = 2\tan^{-1}(x) \), prove that \( x = \frac{a+b}{1-ab} \) Solution: Use the identity: \[ \sin^{-1}\left(\frac{2t}{1+t^2}\right) = 2\tan^{-1}(t) \] Thus, \[ \sin^{-1}\left(\frac{2a}{1+a^2}\right) = 2\tan^{-1}(a) \] \[ \sin^{-1}\left(\frac{2b}{1+b^2}\right) = 2\tan^{-1}(b) \] So the given equation becomes: \[ 2\tan^{-1}(a) + 2\tan^{-1}(b) = 2\tan^{-1}(x) \] Divide both sides by

Prove sin{tan⁻¹((1−x²)/2x) + cot⁻¹((1−x²)/(1+x²))} = 1 Prove that \( \sin\left[ \tan^{-1}\left(\frac{1-x^2}{2x}\right) + \cot^{-1}\left(\frac{1-x^2}{1+x^2}\right) \right] = 1 \) Solution: Let \[ A = \tan^{-1}\left(\frac{1-x^2}{2x}\right) \] \[ B = \cot^{-1}\left(\frac{1-x^2}{1+x^2}\right) \] Convert \(B\) into tangent form: \[ \cot^{-1}(t) = \tan^{-1}\left(\frac{1}{t}\right) \] \[ \Rightarrow B = \tan^{-1}\left(\frac{1+x^2}{1-x^2}\right) \] Now, \[ A + B = \tan^{-1}\left(\frac{1-x^2}{2x}\right) + \tan^{-1}\left(\frac{1+x^2}{1-x^2}\right) \]

Prove tan⁻¹((1−x²)/2x) + cot⁻¹((1−x²)/2x) = π/2 Prove that \( \tan^{-1}\left(\frac{1-x^2}{2x}\right) + \cot^{-1}\left(\frac{1-x^2}{2x}\right) = \frac{\pi}{2} \) Solution: Let \[ \theta = \tan^{-1}\left(\frac{1-x^2}{2x}\right) \] Then, \[ \tan \theta = \frac{1-x^2}{2x} \] Now recall the identity: \[ \cot^{-1}(t) = \tan^{-1}\left(\frac{1}{t}\right) \quad \text{(for } t > 0\text{)} \] So, \[ \cot^{-1}\left(\frac{1-x^2}{2x}\right) = \tan^{-1}\left(\frac{2x}{1-x^2}\right) \] Thus the given expression becomes:

Prove x = (a−b)/(1+ab) If \( \sin^{-1}\left(\frac{2a}{1+a^2}\right) – \cos^{-1}\left(\frac{1-b^2}{1+b^2}\right) = \tan^{-1}\left(\frac{2x}{1-x^2}\right) \), prove that \( x = \frac{a-b}{1+ab} \) Solution: Use standard inverse trigonometric identities: \[ \sin^{-1}\left(\frac{2a}{1+a^2}\right) = 2\tan^{-1}(a) \] \[ \cos^{-1}\left(\frac{1-b^2}{1+b^2}\right) = 2\tan^{-1}(b) \] So the given equation becomes: \[ 2\tan^{-1}(a) – 2\tan^{-1}(b) = \tan^{-1}\left(\frac{2x}{1-x^2}\right) \] \[ \Rightarrow 2\left[\tan^{-1}(a) – \tan^{-1}(b)\right] = \tan^{-1}\left(\frac{2x}{1-x^2}\right) \]

Prove 4tan⁻¹(1/5) − tan⁻¹(1/239) = π/4 Prove that \( 4\tan^{-1}\left(\frac{1}{5}\right) – \tan^{-1}\left(\frac{1}{239}\right) = \frac{\pi}{4} \) Solution: Let \[ \theta = \tan^{-1}\left(\frac{1}{5}\right) \Rightarrow \tan \theta = \frac{1}{5} \] Step 1: Find \( \tan(2\theta) \) \[ \tan(2\theta) = \frac{2\tan\theta}{1 – \tan^2\theta} = \frac{2/5}{1 – 1/25} = \frac{2/5}{24/25} = \frac{5}{12} \] \[ \Rightarrow 2\theta = \tan^{-1}\left(\frac{5}{12}\right) \] Step

Prove 2tan⁻¹(1/2) + tan⁻¹(1/7) = tan⁻¹(31/17) Prove that \( 2\tan^{-1}\left(\frac{1}{2}\right) + \tan^{-1}\left(\frac{1}{7}\right) = \tan^{-1}\left(\frac{31}{17}\right) \) Solution: Let \[ \theta = \tan^{-1}\left(\frac{1}{2}\right) \] Then, \[ \tan \theta = \frac{1}{2} \] Using double angle identity: \[ \tan(2\theta) = \frac{2\tan\theta}{1 – \tan^2\theta} \] \[ = \frac{2 \cdot \frac{1}{2}}{1 – \frac{1}{4}} = \frac{1}{3/4} = \frac{4}{3} \] \[ \Rightarrow 2\theta

Prove 2tan⁻¹(3/4) − tan⁻¹(17/31) = π/4 Prove that \( 2\tan^{-1}\left(\frac{3}{4}\right) – \tan^{-1}\left(\frac{17}{31}\right) = \frac{\pi}{4} \) Solution: Let \[ \theta = \tan^{-1}\left(\frac{3}{4}\right) \] Then, \[ \tan \theta = \frac{3}{4} \] Using identity: \[ \tan(2\theta) = \frac{2\tan\theta}{1 – \tan^2\theta} \] \[ = \frac{2 \cdot \frac{3}{4}}{1 – \left(\frac{3}{4}\right)^2} = \frac{6/4}{1 – 9/16} = \frac{6/4}{7/16} = \frac{24}{7} \] \[

Prove 2tan⁻¹(1/5) + tan⁻¹(1/8) = tan⁻¹(4/7) Prove that \( 2\tan^{-1}\left(\frac{1}{5}\right) + \tan^{-1}\left(\frac{1}{8}\right) = \tan^{-1}\left(\frac{4}{7}\right) \) Solution: Let \[ \theta = \tan^{-1}\left(\frac{1}{5}\right) \] Then, \[ \tan \theta = \frac{1}{5} \] Using double angle identity: \[ \tan(2\theta) = \frac{2\tan\theta}{1 – \tan^2\theta} \] \[ = \frac{2 \cdot \frac{1}{5}}{1 – \frac{1}{25}} = \frac{2/5}{24/25} = \frac{5}{12} \] \[ \Rightarrow 2\theta