Ravi Kant Kumar

Evaluate sin⁻¹(sin 17π/8) Evaluate \( \sin^{-1}(\sin \frac{17\pi}{8}) \) Step-by-Step Solution We need to evaluate: \[ \sin^{-1}\left(\sin \frac{17\pi}{8}\right) \] Step 1: Use periodic property \[ \frac{17\pi}{8} = 2\pi + \frac{\pi}{8} \] \[ \sin\left(\frac{17\pi}{8}\right) = \sin\left(2\pi + \frac{\pi}{8}\right) = \sin\left(\frac{\pi}{8}\right) \] Step 2: Apply inverse sine \[ \sin^{-1}\left(\sin \frac{\pi}{8}\right) \] The principal value range of \( \sin^{-1}x […]

Evaluate sin⁻¹(sin 13π/7) Evaluate \( \sin^{-1}(\sin \frac{13\pi}{7}) \) Step-by-Step Solution We need to evaluate: \[ \sin^{-1}\left(\sin \frac{13\pi}{7}\right) \] Step 1: Use periodic property \[ \frac{13\pi}{7} = 2\pi – \frac{\pi}{7} \] \[ \sin\left(\frac{13\pi}{7}\right) = \sin\left(2\pi – \frac{\pi}{7}\right) = -\sin\left(\frac{\pi}{7}\right) \] Step 2: Apply inverse sine \[ \sin^{-1}\left(-\sin\frac{\pi}{7}\right) \] The principal value range of \( \sin^{-1}x \)

Evaluate sin⁻¹(sin 5π/6) Evaluate \( \sin^{-1}(\sin \frac{5\pi}{6}) \) Step-by-Step Solution We need to evaluate: \[ \sin^{-1}\left(\sin \frac{5\pi}{6}\right) \] Step 1: Find the value of sine \[ \sin \frac{5\pi}{6} = \frac{1}{2} \] Step 2: Apply inverse sine \[ \sin^{-1}\left(\frac{1}{2}\right) \] The principal value range of \( \sin^{-1}x \) is: \[ \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \] Since \( \frac{\pi}{6}

Evaluate sin⁻¹(sin 7π/6) Evaluate \( \sin^{-1}(\sin \frac{7\pi}{6}) \) Step-by-Step Solution We need to evaluate: \[ \sin^{-1}\left(\sin \frac{7\pi}{6}\right) \] Step 1: Find the value of sine \[ \sin \frac{7\pi}{6} = -\frac{1}{2} \] Step 2: Apply inverse sine \[ \sin^{-1}\left(-\frac{1}{2}\right) \] The principal value range of \( \sin^{-1}x \) is: \[ \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \] Since the angle

Evaluate sin⁻¹(sin π/6) Evaluate \( \sin^{-1}(\sin \frac{\pi}{6}) \) Step-by-Step Solution We need to evaluate: \[ \sin^{-1}\left(\sin \frac{\pi}{6}\right) \] Step 1: Find the value of sine \[ \sin \frac{\pi}{6} = \frac{1}{2} \] Step 2: Apply inverse sine \[ \sin^{-1}\left(\frac{1}{2}\right) \] The principal value range of \( \sin^{-1}x \) is: \[ \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \] Since \( \frac{\pi}{6}

Principal Value of tan⁻¹(−1/√3) + cot⁻¹(1/√3) + tan⁻¹(sin(−π/2)) Evaluate: tan-1(−1/√3) + cot-1(1/√3) + tan-1(sin(−π/2)) Solution: Step 1: Evaluate tan⁻¹(−1/√3) \[ \tan^{-1}\left(-\frac{1}{\sqrt{3}}\right) = -\frac{\pi}{6} \] Step 2: Evaluate cot⁻¹(1/√3) \[ \cot^{-1}\left(\frac{1}{\sqrt{3}}\right) = \frac{\pi}{3} \] Step 3: Evaluate sin(−π/2) \[ \sin\left(-\frac{\pi}{2}\right) = -1 \] Step 4: Evaluate tan⁻¹(−1) \[ \tan^{-1}(-1) = -\frac{\pi}{4} \] Step 5: Add all

Principal Value of cosec⁻¹(−2/√3) + 2cot⁻¹(−1) Evaluate: cosec-1(−2/√3) + 2cot-1(−1) Solution: Step 1: Evaluate cosec⁻¹(−2/√3) \[ \csc^{-1}\left(-\frac{2}{\sqrt{3}}\right) \Rightarrow \sin y = -\frac{\sqrt{3}}{2} \] \[ y = -\frac{\pi}{3} \] Step 2: Evaluate cot⁻¹(−1) \[ \cot^{-1}(-1) \] We know: \[ \cot\left(\frac{3\pi}{4}\right) = -1 \] \[ \Rightarrow \cot^{-1}(-1) = \frac{3\pi}{4} \] Step 3: Substitute \[ -\frac{\pi}{3} + 2\left(\frac{3\pi}{4}\right)

Principal Value of cot⁻¹{2cos(sin⁻¹(√3/2))} Evaluate: cot-1{2cos(sin-1(√3/2))} Solution: Step 1: Evaluate sin⁻¹(√3/2) \[ \sin^{-1}\left(\frac{\sqrt{3}}{2}\right) = \frac{\pi}{3} \] Step 2: Substitute \[ 2\cos\left(\frac{\pi}{3}\right) \] \[ = 2 \times \frac{1}{2} = 1 \] Step 3: Apply cot⁻¹ \[ \cot^{-1}(1) \] We know: \[ \cot\left(\frac{\pi}{4}\right) = 1 \] Principal value range of cot⁻¹(x): \[ (0, \pi) \] So, \[

Principal Value of cot⁻¹(1/√3) − cosec⁻¹(−2) + sec⁻¹(2/√3) Evaluate: cot-1(1/√3) − cosec-1(−2) + sec-1(2/√3) Solution: Step 1: Evaluate cot⁻¹(1/√3) \[ \cot^{-1}\left(\frac{1}{\sqrt{3}}\right) = \frac{\pi}{3} \] Step 2: Evaluate cosec⁻¹(−2) \[ \csc^{-1}(-2) = -\frac{\pi}{6} \] Step 3: Evaluate sec⁻¹(2/√3) \[ \sec^{-1}\left(\frac{2}{\sqrt{3}}\right) = \frac{\pi}{6} \] Step 4: Substitute \[ \frac{\pi}{3} – \left(-\frac{\pi}{6}\right) + \frac{\pi}{6} \] \[ = \frac{\pi}{3}

Domain of cot x + cot⁻¹(x) Find the Domain of f(x) = cot x + cot-1(x) Solution: Given function: \[ f(x) = \cot x + \cot^{-1}(x) \] Step 1: Domain of cot x \[ \cot x = \frac{\cos x}{\sin x} \] So, \[ \sin x \neq 0 \Rightarrow x \neq n\pi,\; n \in \mathbb{Z} \]