Ravi Kant Kumar

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

Principal Value of tan⁻¹(1) + cos⁻¹(−1/2) Question Find the principal value of: \[ \tan^{-1}(1) + \cos^{-1}\left(-\frac{1}{2}\right) \] Solution Using standard values: \[ \tan^{-1}(1) = \frac{\pi}{4} \] And, \[ \cos^{-1}\left(-\frac{1}{2}\right) = \frac{2\pi}{3} \] (since principal range of \( \cos^{-1}x \) is \( [0, \pi] \)) Therefore, \[ \frac{\pi}{4} + \frac{2\pi}{3} \] Taking LCM: \[ = \frac{3\pi

Value of tan(2tan⁻¹(1/5)) Question Evaluate: \[ \tan\left(2\tan^{-1}\left(\frac{1}{5}\right)\right) \] Solution Let \[ \theta = \tan^{-1}\left(\frac{1}{5}\right) \Rightarrow \tan \theta = \frac{1}{5} \] Using identity: \[ \tan 2\theta = \frac{2\tan \theta}{1 – \tan^2 \theta} \] Substitute: \[ \tan 2\theta = \frac{2 \cdot \frac{1}{5}}{1 – \left(\frac{1}{5}\right)^2} \] \[ = \frac{2/5}{1 – 1/25} = \frac{2/5}{24/25} \] \[ = \frac{2}{5} \cdot

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

Principal Value of sin⁻¹(−1/2) Question Find the principal value of: \[ \sin^{-1}\left(-\frac{1}{2}\right) \] Solution The principal value range of \( \sin^{-1}x \) is: \[ \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \] We know: \[ \sin \frac{\pi}{6} = \frac{1}{2} \] So, \[ \sin^{-1}\left(-\frac{1}{2}\right) = -\frac{\pi}{6} \] (since \( -\frac{\pi}{6} \) lies in the principal range) Final Answer: \[ \boxed{-\frac{\pi}{6}} \]

Principal Value of sin⁻¹(−√3/2) Question Find the principal value of: \[ \sin^{-1}\left(-\frac{\sqrt{3}}{2}\right) \] Solution The principal value range of \( \sin^{-1}x \) is: \[ \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \] We know: \[ \sin \frac{\pi}{3} = \frac{\sqrt{3}}{2} \] So, \[ \sin^{-1}\left(-\frac{\sqrt{3}}{2}\right) = -\frac{\pi}{3} \] (since \( -\frac{\pi}{3} \) lies in the principal range) Final Answer: \[ \boxed{-\frac{\pi}{3}} \]

If x < 0, y < 0 and xy = 1, find tan⁻¹x + tan⁻¹y Question If \( x < 0 \), \( y < 0 \) and \( xy = 1 \), find: \[ \tan^{-1}x + \tan^{-1}y \] Solution We use identity: \[ \tan^{-1}x + \tan^{-1}y = \tan^{-1}\left(\frac{x + y}{1 – xy}\right) \] Given: