Ravi Kant Kumar

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} –

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, \[ 2\tan \frac{3\pi}{4} = 2 \times (-1) = -2 \] Step 2: Convert to cosine \[ \sec y = -2 \Rightarrow \cos y =

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, \[ 2\sin \frac{3\pi}{4} = 2 \times \frac{1}{\sqrt{2}} = \sqrt{2} \] Step 2: Convert to cosine \[ \sec y = \sqrt{2} \Rightarrow \cos y =

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: \[ \cos\left(\frac{\pi}{3}\right) = \frac{1}{2} \] Since the principal value range of sec-1(x) is: \[ [0,\pi], \quad y \ne \frac{\pi}{2} \] Therefore,

Principal Value of sec⁻¹(−√2) Find the Principal Value of sec-1(−√2) Solution: Let \[ y = \sec^{-1}(-\sqrt{2}) \] Then, \[ \sec y = -\sqrt{2} \] Taking reciprocal: \[ \cos y = -\frac{1}{\sqrt{2}} \] We know: \[ \cos\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}} \] So, \[ \cos y = \cos\left(\pi – \frac{\pi}{4}\right) = \cos\left(\frac{3\pi}{4}\right) \] Since principal value range of

Principal Value of tan⁻¹(tan 5π/6) + cos⁻¹(cos 13π/6) Evaluate: tan-1(tan 5π/6) + cos-1(cos 13π/6) Solution: Step 1: Use principal value ranges tan-1(x) ∈ \((-π/2, π/2)\) cos-1(x) ∈ \([0, π]\) Step 2: Evaluate tan⁻¹(tan 5π/6) \[ \tan \frac{5\pi}{6} = -\tan \frac{\pi}{6} = -\frac{1}{\sqrt{3}} \] \[ \tan^{-1}\left(-\frac{1}{\sqrt{3}}\right) = -\frac{\pi}{6} \] Step 3: Evaluate cos⁻¹(cos 13π/6) \[ \cos

Principal Value of tan⁻¹(−1/√3) + tan⁻¹(−√3) + tan⁻¹(sin(−π/2)) Evaluate: tan-1(−1/√3) + tan-1(−√3) + tan-1(sin(−π/2)) Solution: Using known values: \[ \tan^{-1}\left(-\frac{1}{\sqrt{3}}\right) = -\frac{\pi}{6} \] \[ \tan^{-1}(-\sqrt{3}) = -\frac{\pi}{3} \] \[ \sin(-\frac{\pi}{2}) = -1 \] \[ \tan^{-1}(-1) = -\frac{\pi}{4} \] (Using standard principal values of inverse trigonometric functions :contentReference[oaicite:0]{index=0}) Now, \[ -\frac{\pi}{6} – \frac{\pi}{3} – \frac{\pi}{4} \]

Principal Value of tan⁻¹(1) + cos⁻¹(−1/2) + sin⁻¹(−1/2) Evaluate: tan-1(1) + cos-1(−1/2) + sin-1(−1/2) Solution: Using principal values: \[ \tan^{-1}(1) = \frac{\pi}{4} \] \[ \cos^{-1}\left(-\frac{1}{2}\right) = \frac{2\pi}{3} \] \[ \sin^{-1}\left(-\frac{1}{2}\right) = -\frac{\pi}{6} \] (Using standard principal value ranges of inverse trigonometric functions :contentReference[oaicite:0]{index=0}) Now, \[ \frac{\pi}{4} + \frac{2\pi}{3} – \frac{\pi}{6} \] Take LCM = 12: