Ravi Kant Kumar

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 that: \[ \csc^{-1}(x) = \sin^{-1}\left(\frac{1}{x}\right) \] So, \[ \csc^{-1}\left(\frac{\sqrt{3}}{2}\right) = \sin^{-1}\left(\frac{2}{\sqrt{3}}\right) \] But, \[ \frac{2}{\sqrt{3}} > 1 \] And we know: \[ -1 \le \sin \theta \le 1 \] So, \[ \sin^{-1}\left(\frac{2}{\sqrt{3}}\right) \text{ is not defined in […]

Principal Value of sin⁻¹{cos(sin⁻¹(1/2))} Question Find the principal value of: \[ \sin^{-1}\left\{\cos\left(\sin^{-1}\left(\frac{1}{2}\right)\right)\right\} \] Solution First, evaluate: \[ \sin^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{6} \] So, \[ \cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2} \] Now evaluate: \[ \sin^{-1}\left(\frac{\sqrt{3}}{2}\right) \] The principal value range of \( \sin^{-1}x \) is: \[ \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \] Since \( \frac{\sqrt{3}}{2} \) corresponds to \( \frac{\pi}{3} \) in

Value of tan((sin⁻¹x + cos⁻¹x)/2) when x = √3/2 Question Evaluate: \[ \tan\left(\frac{\sin^{-1}x + \cos^{-1}x}{2}\right) \quad \text{when } x = \frac{\sqrt{3}}{2} \] Solution We use identity: \[ \sin^{-1}x + \cos^{-1}x = \frac{\pi}{2} \] So, \[ \tan\left(\frac{\sin^{-1}x + \cos^{-1}x}{2}\right) = \tan\left(\frac{\pi/2}{2}\right) \] \[ = \tan\left(\frac{\pi}{4}\right) \] \[ = 1 \] Final Answer: \[ \boxed{1} \] Key

Value of cos(sin⁻¹x + cos⁻¹x) Question Find the value of: \[ \cos(\sin^{-1}x + \cos^{-1}x), \quad |x| \le 1 \] Solution We use identity: \[ \sin^{-1}x + \cos^{-1}x = \frac{\pi}{2} \quad \text{for } |x| \le 1 \] So, \[ \cos(\sin^{-1}x + \cos^{-1}x) = \cos\left(\frac{\pi}{2}\right) \] \[ = 0 \] Final Answer: \[ \boxed{0} \] Key Concept

Value of cos⁻¹(cos 14π/3) Question Find the value of: \[ \cos^{-1}(\cos \tfrac{14\pi}{3}) \] Solution First, reduce the angle using periodicity: \[ \frac{14\pi}{3} = 4\pi + \frac{2\pi}{3} \Rightarrow \cos \tfrac{14\pi}{3} = \cos \tfrac{2\pi}{3} \] Now evaluate: \[ \cos^{-1}(\cos \tfrac{2\pi}{3}) \] The principal value range of \( \cos^{-1}x \) is: \[ [0, \pi] \] Since \( \tfrac{2\pi}{3}

Value of sec⁻¹(1/2) Question Find the value of: \[ \sec^{-1}\left(\frac{1}{2}\right) \] Solution We know: \[ \sec^{-1}(x) = \cos^{-1}\left(\frac{1}{x}\right) \] So, \[ \sec^{-1}\left(\frac{1}{2}\right) = \cos^{-1}(2) \] But cosine function satisfies: \[ -1 \le \cos \theta \le 1 \] Since \( 2 \) is outside this range, \[ \cos^{-1}(2) \text{ is not defined in real numbers} \]

Value of sin⁻¹(sin 3π/5) Question Evaluate: \[ \sin^{-1}(\sin \tfrac{3\pi}{5}) \] Solution The principal value range of \( \sin^{-1}x \) is: \[ \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \] Now, \[ \frac{3\pi}{5} \in \left(\frac{\pi}{2}, \pi\right) \] So we use identity: \[ \sin^{-1}(\sin x) = \pi – x \quad \text{for } \frac{\pi}{2} < x < \pi \] Thus, \[ \sin^{-1}(\sin \tfrac{3\pi}{5})

Principal Value of cos⁻¹(cos 680°) Question Find the principal value of: \[ \cos^{-1}(\cos 680^\circ) \] Solution First, reduce the angle: \[ 680^\circ = 360^\circ + 320^\circ \Rightarrow \cos 680^\circ = \cos 320^\circ \] Now, \[ 320^\circ = 360^\circ – 40^\circ \Rightarrow \cos 320^\circ = \cos 40^\circ \] So, \[ \cos^{-1}(\cos 680^\circ) = \cos^{-1}(\cos 40^\circ) \]

Principal Value of tan⁻¹√3 + cot⁻¹√3 Question Find the principal value of: \[ \tan^{-1}(\sqrt{3}) + \cot^{-1}(\sqrt{3}) \] Solution Using standard values: \[ \tan^{-1}(\sqrt{3}) = \frac{\pi}{3} \] Also, \[ \cot^{-1}(\sqrt{3}) = \frac{\pi}{6} \] (since \( \cot \frac{\pi}{6} = \sqrt{3} \)) Therefore, \[ \frac{\pi}{3} + \frac{\pi}{6} = \frac{\pi}{2} \] Final Answer: \[ \boxed{\frac{\pi}{2}} \] Key Concept Use

Value of tan⁻¹{2sin(2cos⁻¹(√3/2))} Question Evaluate: \[ \tan^{-1}\left\{2\sin\left(2\cos^{-1}\left(\frac{\sqrt{3}}{2}\right)\right)\right\} \] Solution First, evaluate: \[ \cos^{-1}\left(\frac{\sqrt{3}}{2}\right) = \frac{\pi}{6} \] So, \[ 2\cos^{-1}\left(\frac{\sqrt{3}}{2}\right) = \frac{\pi}{3} \] Now, \[ \sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2} \] Thus, \[ 2 \cdot \frac{\sqrt{3}}{2} = \sqrt{3} \] Now evaluate: \[ \tan^{-1}(\sqrt{3}) = \frac{\pi}{3} \] Final Answer: \[ \boxed{\frac{\pi}{3}} \] Key Concept Break the expression step-by-step using