Ravi Kant Kumar

Find x² from tan⁻¹ expression Question If \[ \tan^{-1}\left(\frac{\sqrt{1+x^2} – \sqrt{1-x^2}}{\sqrt{1+x^2} + \sqrt{1-x^2}}\right) = \alpha \] Find \( x^2 \). Solution Let \[ \tan^{-1}(A) = \alpha \Rightarrow A = \tan \alpha \] So, \[ \tan \alpha = \frac{\sqrt{1+x^2} – \sqrt{1-x^2}}{\sqrt{1+x^2} + \sqrt{1-x^2}} \] This is a standard identity: \[ \tan\left(\frac{\theta}{2}\right) = \frac{1 – \cos \theta}{\sin […]

Value of tan⁻¹(tan 9π/8) Question Find the value of: \[ \tan^{-1}(\tan \tfrac{9\pi}{8}) \] Solution The principal value range of \( \tan^{-1}x \) is: \[ \left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \] Now reduce the angle: \[ \frac{9\pi}{8} = \pi + \frac{\pi}{8} \Rightarrow \tan \tfrac{9\pi}{8} = \tan \tfrac{\pi}{8} \] But \( \tfrac{\pi}{8} \in \left(0, \frac{\pi}{2}\right) \), so \[ \tan^{-1}(\tan \tfrac{\pi}{8})

Value of cos⁻¹(cos 13π/6) Question Find the value of: \[ \cos^{-1}(\cos \tfrac{13\pi}{6}) \] Solution First, reduce the angle: \[ \frac{13\pi}{6} = 2\pi + \frac{\pi}{6} \Rightarrow \cos \tfrac{13\pi}{6} = \cos \tfrac{\pi}{6} \] Now evaluate: \[ \cos^{-1}(\cos \tfrac{\pi}{6}) \] The principal value range of \( \cos^{-1}x \) is: \[ [0, \pi] \] Since \( \tfrac{\pi}{6} \in [0,

If cos(sin⁻¹(2/5) + cos⁻¹x) = 0, find x Question If \[ \cos\left(\sin^{-1}\left(\frac{2}{5}\right) + \cos^{-1}x\right) = 0 \] Find \( x \). Solution We know: \[ \cos \theta = 0 \Rightarrow \theta = \frac{\pi}{2} \quad \text{(principal value)} \] So, \[ \sin^{-1}\left(\frac{2}{5}\right) + \cos^{-1}x = \frac{\pi}{2} \] Using identity: \[ \sin^{-1}a + \cos^{-1}a = \frac{\pi}{2} \] Comparing,

Value of 2sec⁻¹(2) + sin⁻¹(1/2) Question Evaluate: \[ 2\sec^{-1}(2) + \sin^{-1}\left(\frac{1}{2}\right) \] Solution We know: \[ \sec^{-1}(2) = \cos^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{3} \] So, \[ 2\sec^{-1}(2) = 2 \cdot \frac{\pi}{3} = \frac{2\pi}{3} \] Also, \[ \sin^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{6} \] Therefore, \[ \frac{2\pi}{3} + \frac{\pi}{6} \] \[ = \frac{4\pi}{6} + \frac{\pi}{6} = \frac{5\pi}{6} \] Final Answer: \[

Value of cos((tan⁻¹x + cot⁻¹x)/3) when x = −1/√3 Question Evaluate: \[ \cos\left(\frac{\tan^{-1}x + \cot^{-1}x}{3}\right) \quad \text{when } x = -\frac{1}{\sqrt{3}} \] Solution We use identity: \[ \tan^{-1}x + \cot^{-1}x = \frac{\pi}{2} \quad \text{for all } x \in \mathbb{R} \] So, \[ \cos\left(\frac{\pi/2}{3}\right) = \cos\left(\frac{\pi}{6}\right) \] \[ = \frac{\sqrt{3}}{2} \] Final Answer: \[ \boxed{\frac{\sqrt{3}}{2}} \]

If cos(tan⁻¹x + cot⁻¹√3) = 0, find x Question If \[ \cos\left(\tan^{-1}x + \cot^{-1}\sqrt{3}\right) = 0 \] Find \( x \). Solution We know: \[ \cot^{-1}\sqrt{3} = \frac{\pi}{6} \] So the equation becomes: \[ \cos\left(\tan^{-1}x + \frac{\pi}{6}\right) = 0 \] Cosine is zero when: \[ \theta = \frac{\pi}{2} \quad \text{(within principal consideration)} \] Thus, \[

Value of cot⁻¹(−x) in terms of cot⁻¹x Question Express: \[ \cot^{-1}(-x) \] in terms of \( \cot^{-1}x \), where \( x \in \mathbb{R} \). Solution Let \[ \cot^{-1}x = \theta \] Then, \[ x = \cot \theta \] So, \[ -x = -\cot \theta = \cot(\pi – \theta) \] Thus, \[ \cot^{-1}(-x) = \pi –

Value of tan⁻¹(1/x) for x < 0 in terms of cot⁻¹x Question Express the value of: \[ \tan^{-1}\left(\frac{1}{x}\right) \quad \text{for } x < 0 \] in terms of \( \cot^{-1}x \). Solution Let \[ \cot^{-1}x = \theta \] Then, \[ x = \cot \theta \] So, \[ \frac{1}{x} = \tan \theta \] Thus, \[ \tan^{-1}\left(\frac{1}{x}\right)

Set of Values of cosec⁻¹(√3/2) Question Find the set of values of: \[ \csc^{-1}\left(\frac{\sqrt{3}}{2}\right) \] Solution We know the domain of inverse cosecant: \[ |x| \ge 1 \] But, \[ \frac{\sqrt{3}}{2} < 1 \] So, the value lies outside the domain. Hence, \[ \csc^{-1}\left(\frac{\sqrt{3}}{2}\right) \text{ is not defined in real numbers} \] Final Answer: \[