April 2026

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: \[ 2\tan^{-1}\left(\frac{b}{a}\right) + 2\tan^{-1}\left(\frac{y}{x}\right) \]

Prove 2tan⁻¹{√(a−b)/√(a+b) tan(θ/2)} = cos⁻¹{(a cosθ + b)/(a + b cosθ)} Prove that \( 2\tan^{-1}\left(\sqrt{\frac{a-b}{a+b}} \tan\frac{\theta}{2}\right) = \cos^{-1}\left(\frac{a\cos\theta + b}{a + b\cos\theta}\right) \) Solution: Let \[ t = \sqrt{\frac{a-b}{a+b}} \tan\frac{\theta}{2} \] Then LHS becomes: \[ 2\tan^{-1}(t) \] Use identity: \[ \cos(2\tan^{-1}t) = \frac{1 – t^2}{1 + t^2} \] So, \[ \cos(\text{LHS}) = \frac{1 – t^2}{1

Prove 2tan⁻¹((x−2)/(x−1)) + tan⁻¹((x+2)/(x+1)) = π/4 Prove that \( 2\tan^{-1}\left(\frac{x-2}{x-1}\right) + \tan^{-1}\left(\frac{x+2}{x+1}\right) = \frac{\pi}{4} \) Solution: Let \[ \theta = \tan^{-1}\left(\frac{x-2}{x-1}\right) \Rightarrow \tan\theta = \frac{x-2}{x-1} \] Using identity: \[ \tan(2\theta) = \frac{2\tan\theta}{1 – \tan^2\theta} \] \[ = \frac{2\cdot\frac{x-2}{x-1}}{1 – \left(\frac{x-2}{x-1}\right)^2} \] \[ = \frac{2(x-2)(x-1)}{(x-1)^2 – (x-2)^2} \] \[ = \frac{2(x-2)(x-1)}{(x^2 -2x +1) – (x^2 -4x

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)