Ravi Kant Kumar

Principal Value of tan⁻¹{2sin(4cos⁻¹(√3/2))} Evaluate: tan-1{2sin(4cos-1(√3/2))} Solution: Given: \[ y = \tan^{-1}\left(2\sin\left(4\cos^{-1}\frac{\sqrt{3}}{2}\right)\right) \] Step 1: Evaluate cos-1(√3/2) \[ \cos^{-1}\left(\frac{\sqrt{3}}{2}\right) = \frac{\pi}{6} \] Step 2: Substitute \[ y = \tan^{-1}\left(2\sin\left(4 \cdot \frac{\pi}{6}\right)\right) = \tan^{-1}\left(2\sin\left(\frac{2\pi}{3}\right)\right) \] Step 3: Evaluate sin(2π/3) \[ \sin\frac{2\pi}{3} = \frac{\sqrt{3}}{2} \] So, \[ 2 \times \frac{\sqrt{3}}{2} = \sqrt{3} \] Step 4: Final evaluation […]

Principal Value of tan⁻¹(−1) + cos⁻¹(−1/√2) Evaluate: tan-1(−1) + cos-1(−1/√2) Solution: Using principal values: \[ \tan^{-1}(-1) = -\frac{\pi}{4} \] (Range of tan-1(x) is \((- \pi/2, \pi/2)\)) \[ \cos^{-1}\left(-\frac{1}{\sqrt{2}}\right) = \frac{3\pi}{4} \] (Range of cos-1(x) is \([0, \pi]\)) Now, \[ \tan^{-1}(-1) + \cos^{-1}\left(-\frac{1}{\sqrt{2}}\right) = -\frac{\pi}{4} + \frac{3\pi}{4} \] \[ = \frac{2\pi}{4} = \frac{\pi}{2} \] Final Answer:

Principal Value of tan⁻¹(2cos 2π/3) Find the Principal Value of tan-1(2cos 2π/3) Solution: Given: \[ y = \tan^{-1}\left(2\cos \frac{2\pi}{3}\right) \] Step 1: Evaluate cos(2π/3) \[ \cos \frac{2\pi}{3} = -\frac{1}{2} \] So, \[ 2\cos \frac{2\pi}{3} = 2 \times \left(-\frac{1}{2}\right) = -1 \] Step 2: Use inverse tangent \[ y = \tan^{-1}(-1) \] We know: \[ \tan^{-1}(-1)

Principal Value of tan⁻¹(cos π/2) Find the Principal Value of tan-1(cos π/2) Solution: Given: \[ y = \tan^{-1}(\cos \tfrac{\pi}{2}) \] Step 1: Evaluate cos(π/2) \[ \cos \tfrac{\pi}{2} = 0 \] So, \[ y = \tan^{-1}(0) \] Step 2: Use principal value We know: \[ \tan^{-1}(0) = 0 \] (Principal value range of tan-1(x) is \((-

Principal Value of tan⁻¹(1/√3) Find the Principal Value of tan-1(1/√3) Solution: Let \[ y = \tan^{-1}\left(\frac{1}{\sqrt{3}}\right) \] Then, \[ \tan y = \frac{1}{\sqrt{3}} \] We know that: \[ \tan\left(\frac{\pi}{6}\right) = \frac{1}{\sqrt{3}} \] Since the principal value range of tan-1(x) is: \[ \left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \] Therefore, \[ y = \frac{\pi}{6} \] Final Answer: Principal Value =

Principal Value of tan⁻¹(−1/√3) Find the Principal Value of tan-1(−1/√3) Solution: Let \[ y = \tan^{-1}\left(-\frac{1}{\sqrt{3}}\right) \] Then, \[ \tan y = -\frac{1}{\sqrt{3}} \] We know: \[ \tan\left(\frac{\pi}{6}\right) = \frac{1}{\sqrt{3}} \] Using the identity: \[ \tan^{-1}(-x) = -\tan^{-1}(x) \] So, \[ y = -\frac{\pi}{6} \] Since the principal value range of tan-1(x) is: \[ \left(-\frac{\pi}{2},

Principal Value of sin⁻¹(−√3/2) + cos⁻¹(√3/2) Evaluate: sin-1(−√3/2) + cos-1(√3/2) Solution: Using principal values: \[ \sin^{-1}\left(-\frac{\sqrt{3}}{2}\right) = -\frac{\pi}{3} \] (Since range of sin-1(x) is \([- \pi/2, \pi/2]\)) \[ \cos^{-1}\left(\frac{\sqrt{3}}{2}\right) = \frac{\pi}{6} \] (Since range of cos-1(x) is \([0, \pi]\)) Substitute values: \[ -\frac{\pi}{3} + \frac{\pi}{6} = -\frac{2\pi}{6} + \frac{\pi}{6} = -\frac{\pi}{6} \] Final Answer: Value

Principal Value of sin⁻¹(−1/2) + 2cos⁻¹(−√3/2) Evaluate: sin-1(−1/2) + 2cos-1(−√3/2) Solution: Using principal values: \[ \sin^{-1}\left(-\frac{1}{2}\right) = -\frac{\pi}{6} \] (Since range of sin-1(x) is \([- \pi/2, \pi/2]\)) \[ \cos^{-1}\left(-\frac{\sqrt{3}}{2}\right) = \frac{5\pi}{6} \] (Since range of cos-1(x) is \([0, \pi]\)) Substitute values: \[ \sin^{-1}\left(-\frac{1}{2}\right) + 2\cos^{-1}\left(-\frac{\sqrt{3}}{2}\right) = -\frac{\pi}{6} + 2 \times \frac{5\pi}{6} \] \[ = -\frac{\pi}{6}

Principal Value of cos⁻¹(1/2) − 2sin⁻¹(−1/2) Evaluate: cos-1(1/2) − 2sin-1(−1/2) Solution: We know the principal values: \[ \cos^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{3} \] \[ \sin^{-1}\left(-\frac{1}{2}\right) = -\frac{\pi}{6} \] (Since the range of sin-1(x) is \([- \pi/2, \pi/2]\)) Substitute values: \[ \cos^{-1}\left(\frac{1}{2}\right) – 2\sin^{-1}\left(-\frac{1}{2}\right) = \frac{\pi}{3} – 2\left(-\frac{\pi}{6}\right) \] \[ = \frac{\pi}{3} + \frac{\pi}{3} = \frac{2\pi}{3} \] Final

Principal Value of cos⁻¹(1/2) + 2sin⁻¹(1/2) Evaluate: cos-1(1/2) + 2sin-1(1/2) Solution: We know: \[ \cos^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{3} \] \[ \sin^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{6} \] Substitute values: \[ \cos^{-1}\left(\frac{1}{2}\right) + 2\sin^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{3} + 2 \times \frac{\pi}{6} \] \[ = \frac{\pi}{3} + \frac{\pi}{3} = \frac{2\pi}{3} \] Final Answer: Value = \[ \frac{2\pi}{3} \] Next Question / Full