Ravi Kant Kumar

Range of tan⁻¹x Question Write the range of: \[ \tan^{-1}x \] Solution The inverse tangent function \( \tan^{-1}x \) is defined for all real values of \( x \). Its principal value range is: \[ \left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \] This means: \( \tan^{-1}x \) never equals \( \frac{\pi}{2} \) or \( -\frac{\pi}{2} \) But it approaches […]

Value of cos⁻¹(1/2) + 2sin⁻¹(1/2) Question Find the value of: \[ \cos^{-1}\left(\frac{1}{2}\right) + 2\sin^{-1}\left(\frac{1}{2}\right) \] Solution Using standard values: \[ \cos^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{3} \] (since \( \cos \frac{\pi}{3} = \frac{1}{2} \) and lies in principal range) Also, \[ \sin^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{6} \] (since \( \sin \frac{\pi}{6} = \frac{1}{2} \)) Therefore, \[ \cos^{-1}\left(\frac{1}{2}\right) + 2\sin^{-1}\left(\frac{1}{2}\right) =

Value of sin(cot⁻¹x) Question Find the value of: \[ \sin(\cot^{-1}x) \] Solution Let \[ \theta = \cot^{-1}x \] Then, \[ \cot \theta = x = \frac{\text{Adjacent}}{\text{Opposite}} \] Take a right triangle such that: Adjacent = \( x \) Opposite = \( 1 \) Then hypotenuse: \[ \sqrt{x^2 + 1} \] Now, \[ \sin \theta =

If −1 < x < 0, find sin⁻¹(2x/(1+x²)) + cos⁻¹((1−x²)/(1+x²)) Question If \( -1 < x < 0 \), find the value of: \[ \sin^{-1}\left(\frac{2x}{1+x^2}\right) + \cos^{-1}\left(\frac{1-x^2}{1+x^2}\right) \] Solution Let \[ x = \tan \theta \] Then, since \( -1 < x < 0 \Rightarrow \theta \in \left(-\frac{\pi}{4}, 0\right) \) Using identities: \[ \sin 2\theta

Value of cos⁻¹(cos 2π/3) + sin⁻¹(sin 2π/3) Question Find the value of: \[ \cos^{-1}(\cos \tfrac{2\pi}{3}) + \sin^{-1}(\sin \tfrac{2\pi}{3}) \] Solution We use principal value ranges: \( \cos^{-1}x \in [0, \pi] \) \( \sin^{-1}x \in [-\frac{\pi}{2}, \frac{\pi}{2}] \) Now, \[ \cos^{-1}(\cos \tfrac{2\pi}{3}) = \tfrac{2\pi}{3} \] Since \( \tfrac{2\pi}{3} \in [0, \pi] \) Next, \[ \sin^{-1}(\sin \tfrac{2\pi}{3})

Value of tan⁻¹x + tan⁻¹(1/x) for x < 0 Question Find the value of: \[ \tan^{-1}x + \tan^{-1}\left(\frac{1}{x}\right) \] given that \( x < 0 \). Solution Let \[ \tan^{-1}x = \theta \] Then, \[ x = \tan \theta \] Since \( x < 0 \Rightarrow \theta < 0 \). Now, \[ \tan^{-1}\left(\frac{1}{x}\right) = \tan^{-1}(\cot\theta)

Value of tan⁻¹x + tan⁻¹(1/x) for x > 0 Question Find the value of: \[ \tan^{-1}x + \tan^{-1}\left(\frac{1}{x}\right) \] given that \( x > 0 \). Solution Let \[ \tan^{-1}x = \theta \] Then, \[ x = \tan \theta \] Now, \[ \tan^{-1}\left(\frac{1}{x}\right) = \tan^{-1}\left(\frac{1}{\tan\theta}\right) = \tan^{-1}(\cot\theta) \] We know that: \[ \cot\theta = \tan\left(\frac{\pi}{2}

If x < 0, find cos⁻¹((1−x²)/(1+x²)) in terms of tan⁻¹x Problem If \( x < 0 \), then express: \[ \cos^{-1}\left(\frac{1 – x^2}{1 + x^2}\right) \] in terms of \( \tan^{-1}x \). Solution Let \[ \tan^{-1}x = \theta \] Then, \[ x = \tan \theta \] Using identity: \[ \cos 2\theta = \frac{1 – \tan^2\theta}{1

If x > 1, find sin⁻¹(2x/(1+x²)) in terms of tan⁻¹x Problem If \( x > 1 \), then express: \[ \sin^{-1}\left(\frac{2x}{1+x^2}\right) \] in terms of \( \tan^{-1}x \). Solution Let \[ \tan^{-1}x = \theta \] Then, \[ x = \tan \theta \] Using identity: \[ \sin 2\theta = \frac{2\tan\theta}{1+\tan^2\theta} \] Substitute \( x = \tan

Find x+y+z when sin⁻¹x + sin⁻¹y + sin⁻¹z = 3π/2 Find the Value of x + y + z Given: \[ \sin^{-1}x + \sin^{-1}y + \sin^{-1}z = \frac{3\pi}{2} \] Concept Used: The principal value range of inverse sine is: \[ \sin^{-1}t \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \] Maximum value of each term = \( \frac{\pi}{2} \) Step