Ravi Kant Kumar

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

Teacher Solution: Difference Between Max and Min of sin⁻¹x Difference Between Maximum and Minimum Values of sin⁻¹x Given: \[ x \in [-1, 1] \] To Find: Difference between maximum and minimum values of \[ \sin^{-1}x \] Concept Used: The principal value range of inverse sine function is: \[ \sin^{-1}x \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \] Step 1:

“`html Value of sin⁻¹(-√3/2) + cos⁻¹(-1/2) Find the Value of sin⁻¹(-√3/2) + cos⁻¹(-1/2) Given Expression \[ \sin^{-1}\left(-\frac{\sqrt{3}}{2}\right) + \cos^{-1}\left(-\frac{1}{2}\right) \] Step 1: Use Principal Values The principal value range of: \( \sin^{-1}x \in [-\frac{\pi}{2}, \frac{\pi}{2}] \) \( \cos^{-1}x \in [0, \pi] \) Step 2: Evaluate Each Term We know: \[ \sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2} \] So,

Prove Advanced tan⁻¹ Identity Prove that for \(a,b,x,y > 0\), \[ \frac{2}{3}\tan^{-1}\left(\frac{3ab^2 – a^3}{b^3 – 3a^2b}\right) + \frac{2}{3}\tan^{-1}\left(\frac{3xy^2 – x^3}{y^3 – 3x^2y}\right) = \tan^{-1}\left(\frac{2\alpha\beta}{\alpha^2 – \beta^2}\right) \] where \( \alpha = -ax + by,\; \beta = bx + ay \) Solution: Use the identity: \[ \tan(3\theta) = \frac{3\tan\theta – \tan^3\theta}{1 – 3\tan^2\theta} \] This implies:

Prove tan⁻¹ Identity with α=ax−by, β=ay+bx Prove that \( \tan^{-1}\left(\frac{2ab}{a^2-b^2}\right) + \tan^{-1}\left(\frac{2xy}{x^2-y^2}\right) = \tan^{-1}\left(\frac{2\alpha\beta}{\alpha^2-\beta^2}\right) \) where \( \alpha = ax – by,\; \beta = ay + bx \) Solution: Use identity: \[ \tan^{-1}\left(\frac{2t}{1-t^2}\right) = 2\tan^{-1}(t) \] Rewrite: \[ \tan^{-1}\left(\frac{2ab}{a^2-b^2}\right) = 2\tan^{-1}\left(\frac{b}{a}\right) \] \[ \tan^{-1}\left(\frac{2xy}{x^2-y^2}\right) = 2\tan^{-1}\left(\frac{y}{x}\right) \] So LHS becomes: \[ 2\tan^{-1}\left(\frac{b}{a}\right) + 2\tan^{-1}\left(\frac{y}{x}\right) \]