Ravi Kant Kumar

Solve cos⁻¹((x²−1)/(x²+1)) + ½tan⁻¹(2x/(1−x²)) Solve \( \cos^{-1}\left(\frac{x^2 – 1}{x^2 + 1}\right) + \frac{1}{2}\tan^{-1}\left(\frac{2x}{1 – x^2}\right) = \frac{2\pi}{3} \) Solution: Use identities: \[ \cos^{-1}\left(\frac{1 – t^2}{1 + t^2}\right) = 2\tan^{-1}(t) \] Here, \[ \frac{x^2 – 1}{x^2 + 1} = -\frac{1 – x^2}{1 + x^2} \] So, \[ \cos^{-1}\left(\frac{x^2 – 1}{x^2 + 1}\right) = \pi – 2\tan^{-1}(x) […]

Solve 2tan⁻¹(sin x) = tan⁻¹(2sec x) Solve \( 2\tan^{-1}(\sin x) = \tan^{-1}(2\sec x), \quad x \ne \frac{\pi}{2} \) Solution: Use identity: \[ \tan(2\theta) = \frac{2\tan\theta}{1 – \tan^2\theta} \] Let \[ \theta = \tan^{-1}(\sin x) \Rightarrow \tan \theta = \sin x \] Then, \[ \tan(2\theta) = \frac{2\sin x}{1 – \sin^2 x} = \frac{2\sin x}{\cos^2 x} =

Solve tan⁻¹(2x/(1−x²)) + cot⁻¹((1−x²)/2x) = 2π/3 Solve \( \tan^{-1}\left(\frac{2x}{1-x^2}\right) + \cot^{-1}\left(\frac{1-x^2}{2x}\right) = \frac{2\pi}{3}, \quad x > 0 \) Solution: Use identity: \[ \cot^{-1}(t) = \tan^{-1}\left(\frac{1}{t}\right) \] So, \[ \cot^{-1}\left(\frac{1-x^2}{2x}\right) = \tan^{-1}\left(\frac{2x}{1-x^2}\right) \] Thus equation becomes: \[ 2\tan^{-1}\left(\frac{2x}{1-x^2}\right) = \frac{2\pi}{3} \] \[ \Rightarrow \tan^{-1}\left(\frac{2x}{1-x^2}\right) = \frac{\pi}{3} \] Taking tangent: \[ \frac{2x}{1-x^2} = \tan\left(\frac{\pi}{3}\right) = \sqrt{3} \]

Solve 3sin⁻¹(2x/(1+x²)) − 4cos⁻¹((1−x²)/(1+x²)) + 2tan⁻¹(2x/(1−x²)) Solve \( 3\sin^{-1}\left(\frac{2x}{1+x^2}\right) – 4\cos^{-1}\left(\frac{1-x^2}{1+x^2}\right) + 2\tan^{-1}\left(\frac{2x}{1-x^2}\right) = \frac{\pi}{3} \) Solution: Use standard identities: \[ \sin^{-1}\left(\frac{2x}{1+x^2}\right) = 2\tan^{-1}(x) \] \[ \cos^{-1}\left(\frac{1-x^2}{1+x^2}\right) = 2\tan^{-1}(x) \] \[ \tan^{-1}\left(\frac{2x}{1-x^2}\right) = 2\tan^{-1}(x) \] Substitute into equation: \[ 3(2\tan^{-1}x) – 4(2\tan^{-1}x) + 2(2\tan^{-1}x) = \frac{\pi}{3} \] \[ 6\tan^{-1}x – 8\tan^{-1}x + 4\tan^{-1}x = \frac{\pi}{3}

Solve tan⁻¹(1/4) + 2tan⁻¹(1/5) + tan⁻¹(1/6) + tan⁻¹(1/x) = π/4 Solve \( \tan^{-1}\left(\frac{1}{4}\right) + 2\tan^{-1}\left(\frac{1}{5}\right) + \tan^{-1}\left(\frac{1}{6}\right) + \tan^{-1}\left(\frac{1}{x}\right) = \frac{\pi}{4} \) Solution: Step 1: Evaluate \(2\tan^{-1}(1/5)\) \[ \tan(2\theta) = \frac{2\tan\theta}{1 – \tan^2\theta} \Rightarrow \tan(2\tan^{-1}(1/5)) = \frac{2/5}{1 – 1/25} = \frac{2/5}{24/25} = \frac{5}{12} \] \[ \Rightarrow 2\tan^{-1}\left(\frac{1}{5}\right) = \tan^{-1}\left(\frac{5}{12}\right) \] Step 2: Combine first two

Find cos(sec⁻¹x + cosec⁻¹x) Find the value of \( \cos\left(\sec^{-1}x + \csc^{-1}x\right), \quad |x| \ge 1 \) Solution: Let \[ \theta = \sec^{-1}x \Rightarrow \sec \theta = x \Rightarrow \cos \theta = \frac{1}{x} \] \[ \phi = \csc^{-1}x \Rightarrow \csc \phi = x \Rightarrow \sin \phi = \frac{1}{x} \] Now, \[ \sin \theta = \sqrt{1

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, \[ A + B = \tan^{-1}\left(\frac{1-x^2}{2x}\right) + \tan^{-1}\left(\frac{1+x^2}{1-x^2}\right) \]