Educational

Domain of sec⁻¹(3x − 1) Find the Domain of sec-1(3x − 1) Solution: Given function: \[ f(x) = \sec^{-1}(3x – 1) \] Condition for sec⁻¹(x): \[ |x| \geq 1 \quad \text{i.e.,} \quad x \leq -1 \;\text{or}\; x \geq 1 \] Apply this to \(3x – 1\): \[ |3x – 1| \geq 1 \] Case 1: […]

Principal Value of sin⁻¹(−√3/2) − 2sec⁻¹(2tan π/6) Evaluate: sin-1(−√3/2) − 2sec-1(2tan π/6) Solution: Step 1: Evaluate sin⁻¹(−√3/2) \[ \sin^{-1}\left(-\frac{\sqrt{3}}{2}\right) = -\frac{\pi}{3} \] (Range of sin-1(x): \([- \pi/2, \pi/2]\)) Step 2: Evaluate 2tan(π/6) \[ \tan \frac{\pi}{6} = \frac{1}{\sqrt{3}} \Rightarrow 2\tan \frac{\pi}{6} = \frac{2}{\sqrt{3}} \] Step 3: Evaluate sec⁻¹(2/√3) \[ \sec^{-1}\left(\frac{2}{\sqrt{3}}\right) = \frac{\pi}{6} \] (Since sec θ

Principal Value of tan⁻¹(√3) − sec⁻¹(−2) Evaluate: tan-1(√3) − sec-1(−2) Solution: Step 1: Evaluate tan⁻¹(√3) \[ \tan^{-1}(\sqrt{3}) = \frac{\pi}{3} \] (Principal range: \((- \pi/2, \pi/2)\)) Step 2: Evaluate sec⁻¹(−2) \[ \sec^{-1}(-2) = \frac{2\pi}{3} \] (Since sec θ = −2 ⇒ cos θ = −1/2, and θ ∈ [0,π], θ ≠ π/2) Step 3: Subtract \[

Principal Value of sec⁻¹(2tan 3π/4) Find the Principal Value of sec-1(2tan 3π/4) Solution: Given: \[ y = \sec^{-1}\left(2\tan \frac{3\pi}{4}\right) \] Step 1: Evaluate tan(3π/4) \[ \tan \frac{3\pi}{4} = -1 \] So, \[ 2\tan \frac{3\pi}{4} = -2 \] Step 2: Convert to cosine \[ \sec y = -2 \Rightarrow \cos y = -\frac{1}{2} \] Step 3:

Principal Value of sec⁻¹(2sin 3π/4) Find the Principal Value of sec-1(2sin 3π/4) Solution: Given: \[ y = \sec^{-1}\left(2\sin \frac{3\pi}{4}\right) \] Step 1: Evaluate sin(3π/4) \[ \sin \frac{3\pi}{4} = \frac{1}{\sqrt{2}} \] So, \[ 2\sin \frac{3\pi}{4} = \sqrt{2} \] Step 2: Convert to cosine \[ \sec y = \sqrt{2} \Rightarrow \cos y = \frac{1}{\sqrt{2}} \] Step 3:

Principal Value of sec⁻¹(2) Find the Principal Value of sec-1(2) Solution: Let \[ y = \sec^{-1}(2) \] Then, \[ \sec y = 2 \] Taking reciprocal: \[ \cos y = \frac{1}{2} \] We know: \[ \cos\left(\frac{\pi}{3}\right) = \frac{1}{2} \] Principal value range of sec-1(x): \[ [0,\pi], \quad y \ne \frac{\pi}{2} \] Therefore, \[ y =

Domain of sec⁻¹x − tan⁻¹x Find the Domain of sec-1x − tan-1x Solution: Given function: \[ f(x) = \sec^{-1}(x) – \tan^{-1}(x) \] Step 1: Domain of sec⁻¹(x) \[ x \leq -1 \quad \text{or} \quad x \geq 1 \] Step 2: Domain of tan⁻¹(x) \[ x \in \mathbb{R} \] Step 3: Intersection \[ (-\infty, -1] \cup

Domain of sec⁻¹(3x − 1) Find the Domain of sec-1(3x − 1) Solution: Given function: \[ f(x) = \sec^{-1}(3x – 1) \] Condition for sec⁻¹(x): \[ x \leq -1 \quad \text{or} \quad x \geq 1 \] So, \[ 3x – 1 \leq -1 \quad \text{or} \quad 3x – 1 \geq 1 \] Case 1: \[

Principal Value of sin⁻¹(−√3/2) − 2sec⁻¹(2tan π/6) Evaluate: sin-1(−√3/2) − 2sec-1(2tan π/6) Solution: Step 1: Evaluate sin⁻¹(−√3/2) \[ \sin^{-1}\left(-\frac{\sqrt{3}}{2}\right) = -\frac{\pi}{3} \] (Since range of sin-1(x) is \([- \pi/2, \pi/2]\)) Step 2: Evaluate 2tan(π/6) \[ \tan \frac{\pi}{6} = \frac{1}{\sqrt{3}} \Rightarrow 2\tan \frac{\pi}{6} = \frac{2}{\sqrt{3}} \] Step 3: Evaluate sec⁻¹(2/√3) \[ \sec^{-1}\left(\frac{2}{\sqrt{3}}\right) = \frac{\pi}{6} \] (Because

Principal Value of tan⁻¹(√3) − sec⁻¹(−2) Evaluate: tan-1(√3) − sec-1(−2) Solution: Step 1: Evaluate tan⁻¹(√3) \[ \tan^{-1}(\sqrt{3}) = \frac{\pi}{3} \] (Since principal range of tan-1(x) is \((- \pi/2, \pi/2)\)) Step 2: Evaluate sec⁻¹(−2) \[ \sec^{-1}(-2) = \frac{2\pi}{3} \] (Using identity: sec-1(−x) = π − sec-1(x), and sec-1(2) = π/3) Step 3: Subtract \[ \frac{\pi}{3} –