Educational

Principal Value of cot⁻¹(√3) Find the Principal Value of cot-1(√3) Solution: Let \[ y = \cot^{-1}(\sqrt{3}) \] Then, \[ \cot y = \sqrt{3} \] We know: \[ \cot\left(\frac{\pi}{6}\right) = \sqrt{3} \] Principal value range of cot⁻¹(x): \[ (0, \pi) \] Since \( \frac{\pi}{6} \in (0,\pi) \), \[ y = \frac{\pi}{6} \] Final Answer: Principal Value […]

Principal Value of cosec⁻¹(2tan 11π/6) Evaluate: cosec-1(2tan 11π/6) Solution: Step 1: Evaluate tan(11π/6) \[ \tan \frac{11\pi}{6} = -\frac{1}{\sqrt{3}} \] So, \[ 2\tan \frac{11\pi}{6} = -\frac{2}{\sqrt{3}} \] Step 2: Convert to sine \[ \csc y = -\frac{2}{\sqrt{3}} \Rightarrow \sin y = -\frac{\sqrt{3}}{2} \] Step 3: Find principal value \[ \sin y = -\frac{\sqrt{3}}{2} \Rightarrow y =

Principal Value of sin⁻¹[cos{2cosec⁻¹(−2)}] Evaluate: sin-1[cos{2cosec-1(−2)}] Solution: Step 1: Evaluate cosec⁻¹(−2) \[ \csc^{-1}(-2) \Rightarrow \sin y = -\frac{1}{2} \] \[ y = -\frac{\pi}{6} \] (Principal range: \([-\pi/2,0) \cup (0,\pi/2]\)) Step 2: Multiply by 2 \[ 2y = 2\left(-\frac{\pi}{6}\right) = -\frac{\pi}{3} \] Step 3: Evaluate cosine \[ \cos\left(-\frac{\pi}{3}\right) = \frac{1}{2} \] Step 4: Apply sin⁻¹ \[

Principal Value of sec⁻¹(√2) + 2cosec⁻¹(−√2) Evaluate: sec-1(√2) + 2cosec-1(−√2) Solution: Step 1: Evaluate sec⁻¹(√2) \[ \sec^{-1}(\sqrt{2}) \Rightarrow \cos y = \frac{1}{\sqrt{2}} \] \[ y = \frac{\pi}{4} \] (Principal range: \([0,\pi], y \ne \pi/2\)) Step 2: Evaluate cosec⁻¹(−√2) \[ \csc^{-1}(-\sqrt{2}) \Rightarrow \sin y = -\frac{1}{\sqrt{2}} \] \[ y = -\frac{\pi}{4} \] (Principal range: \([-\pi/2,0) \cup

Principal Value of sin⁻¹(−√3/2) + cosec⁻¹(−2/√3) Evaluate: sin-1(−√3/2) + cosec-1(−2/√3) Solution: Step 1: Evaluate sin⁻¹(−√3/2) \[ \sin^{-1}\left(-\frac{\sqrt{3}}{2}\right) = -\frac{\pi}{3} \] (Principal range: \([- \pi/2, \pi/2]\)) Step 2: Evaluate cosec⁻¹(−2/√3) \[ \csc^{-1}\left(-\frac{2}{\sqrt{3}}\right) \Rightarrow \sin y = -\frac{\sqrt{3}}{2} \] \[ \Rightarrow y = -\frac{\pi}{3} \] (Principal range: \([-\pi/2,0) \cup (0,\pi/2]\)) Step 3: Add values \[ -\frac{\pi}{3} +

Set of Values of cosec⁻¹(√3/2) Find the Set of Values of cosec-1(√3/2) Solution: Given: \[ y = \csc^{-1}\left(\frac{\sqrt{3}}{2}\right) \] Domain condition for cosec⁻¹(x): \[ |x| \ge 1 \] Check the value: \[ \frac{\sqrt{3}}{2} < 1 \] This does not satisfy the required condition. Therefore, cosec⁻¹(√3/2) is not defined in real numbers. Final Answer: Set of

Principal Value of cosec⁻¹(2cos 2π/3) Find the Principal Value of cosec-1(2cos 2π/3) Solution: Given: \[ y = \csc^{-1}\left(2\cos \frac{2\pi}{3}\right) \] Step 1: Evaluate cos(2π/3) \[ \cos \frac{2\pi}{3} = -\frac{1}{2} \] So, \[ 2\cos \frac{2\pi}{3} = -1 \] Step 2: Convert to sine \[ \csc y = -1 \Rightarrow \sin y = -1 \] Step 3:

Principal Value of cosec⁻¹(2/√3) Find the Principal Value of cosec-1(2/√3) Solution: Let \[ y = \csc^{-1}\left(\frac{2}{\sqrt{3}}\right) \] Then, \[ \csc y = \frac{2}{\sqrt{3}} \Rightarrow \sin y = \frac{\sqrt{3}}{2} \] We know: \[ \sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2} \] Principal value range of cosec⁻¹(x): \[ \left[-\frac{\pi}{2}, 0\right) \cup \left(0, \frac{\pi}{2}\right] \] Since \( \frac{\pi}{3} \) lies in \(

Principal Value of cosec⁻¹(−2) Find the Principal Value of cosec-1(−2) Solution: Let \[ y = \csc^{-1}(-2) \] Then, \[ \csc y = -2 \Rightarrow \sin y = -\frac{1}{2} \] We know: \[ \sin\left(\frac{\pi}{6}\right) = \frac{1}{2} \] So, \[ \sin y = -\frac{1}{2} = \sin\left(-\frac{\pi}{6}\right) \] Principal value range of cosec⁻¹(x): \[ \left[-\frac{\pi}{2}, 0\right) \cup \left(0,

Principal Value of cosec⁻¹(−√2) Find the Principal Value of cosec-1(−√2) Solution: Let \[ y = \csc^{-1}(-\sqrt{2}) \] Then, \[ \csc y = -\sqrt{2} \Rightarrow \sin y = -\frac{1}{\sqrt{2}} \] We know: \[ \sin\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}} \] So, \[ \sin y = -\frac{1}{\sqrt{2}} = \sin\left(-\frac{\pi}{4}\right) \] Principal value range of cosec⁻¹(x): \[ \left[-\frac{\pi}{2}, 0\right) \cup \left(0,