Ravi Kant Kumar

Domain of sin⁻¹(√(x² − 1)) Question: Find the domain of: \[ f(x) = \sin^{-1}\!\left(\sqrt{x^2 – 1}\right) \] Concept: For \( \sin^{-1}(t) \) to be defined: \[ -1 \leq t \leq 1 \] Also, the square root requires: \[ x^2 – 1 \geq 0 \] — Solution: Step 1: Square root condition \[ x^2 – 1 […]

Domain of sin⁻¹(x) + sin(x) Question: Find the domain of the function: \[ f(x) = \sin^{-1}(x) + \sin(x) \] Concept: – \( \sin(x) \) is defined for all real numbers – \( \sin^{-1}(x) \) is defined only when: \[ -1 \leq x \leq 1 \] — Solution: Since \( \sin(x) \) has no restriction, the

Domain of sin⁻¹(x²) Question: Find the domain of the function: \[ f(x) = \sin^{-1}(x^2) \] Concept: For \( \sin^{-1}(t) \) to be defined: \[ -1 \leq t \leq 1 \] — Solution: Step 1: Apply condition \[ -1 \leq x^2 \leq 1 \] Step 2: Solve inequality Since \( x^2 \geq 0 \), the condition

Principal Value of sin⁻¹{cos(sin⁻¹(√3/2))} Question: Find the principal value of: \[ \sin^{-1}\left\{\cos\left(\sin^{-1}\left(\frac{\sqrt{3}}{2}\right)\right)\right\} \] Solution: Step 1: Evaluate inner inverse \[ \sin^{-1}\left(\frac{\sqrt{3}}{2}\right) = \frac{\pi}{3} \] Step 2: Substitute \[ \sin^{-1}\left(\cos \frac{\pi}{3}\right) \] Step 3: Evaluate cosine \[ \cos \frac{\pi}{3} = \frac{1}{2} \] Step 4: Final inverse \[ \sin^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{6} \] — Final Answer: \[ \boxed{\frac{\pi}{6}}

Principal Value sin⁻¹(1/2) – 2sin⁻¹(1/√2) Question: Find the principal value of: \[ \sin^{-1}\left(\frac{1}{2}\right) – 2\sin^{-1}\left(\frac{1}{\sqrt{2}}\right) \] Solution: Step 1: Use standard values \[ \sin^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{6} \] \[ \sin^{-1}\left(\frac{1}{\sqrt{2}}\right) = \frac{\pi}{4} \] Step 2: Substitute \[ \frac{\pi}{6} – 2\left(\frac{\pi}{4}\right) \] Step 3: Simplify \[ = \frac{\pi}{6} – \frac{\pi}{2} \] \[ = \frac{\pi – 3\pi}{6} =

Principal Value of sin⁻¹(tan 5π/4) Question: Find the principal value of: \[ \sin^{-1}\left(\tan \frac{5\pi}{4}\right) \] Solution: Step 1: Evaluate tangent value \[ \tan \frac{5\pi}{4} = 1 \] Step 2: Substitute \[ \sin^{-1}(1) \] Step 3: Use standard value \[ \sin^{-1}(1) = \frac{\pi}{2} \] Step 4: Check principal range \[ \frac{\pi}{2} \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \] —

Principal Value of sin⁻¹(cos 3π/4) Question: Find the principal value of: \[ \sin^{-1}\left(\cos \frac{3\pi}{4}\right) \] Solution: Step 1: Evaluate cosine value \[ \cos \frac{3\pi}{4} = -\frac{\sqrt{2}}{2} \] Step 2: Substitute \[ \sin^{-1}\left(-\frac{\sqrt{2}}{2}\right) \] Step 3: Use standard value \[ \sin\left(-\frac{\pi}{4}\right) = -\frac{\sqrt{2}}{2} \] Step 4: Check principal range \[ -\frac{\pi}{4} \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \] —

Principal Value of sin⁻¹((√3+1)/(2√2)) Question: Find the principal value of: \[ \sin^{-1}\left(\frac{\sqrt{3}+1}{2\sqrt{2}}\right) \] Solution: Step 1: Use identity Recall: \[ \sin\left(\frac{5\pi}{12}\right) = \frac{\sqrt{6} + \sqrt{2}}{4} \] Now simplify: \[ \frac{\sqrt{3}+1}{2\sqrt{2}} = \frac{\sqrt{6} + \sqrt{2}}{4} \] Step 2: Substitute \[ \sin^{-1}\left(\frac{\sqrt{6} + \sqrt{2}}{4}\right) \] Step 3: Check principal range \[ \frac{5\pi}{12} \notin \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \] Use

Principal Value of sin⁻¹((√3−1)/(2√2)) Question: Find the principal value of: \[ \sin^{-1}\left(\frac{\sqrt{3}-1}{2\sqrt{2}}\right) \] Solution: Step 1: Use identity Recall: \[ \sin\left(\frac{\pi}{12}\right) = \frac{\sqrt{6} – \sqrt{2}}{4} \] Now simplify given expression: \[ \frac{\sqrt{3}-1}{2\sqrt{2}} = \frac{\sqrt{6} – \sqrt{2}}{4} \] Step 2: Substitute \[ \sin^{-1}\left(\frac{\sqrt{6} – \sqrt{2}}{4}\right) \] Step 3: Recognize value \[ = \frac{\pi}{12} \] Step 4:

Principal Value of sin⁻¹(cos 2π/3) Question: Find the principal value of: \[ \sin^{-1}\left(\cos \frac{2\pi}{3}\right) \] Solution: Step 1: Evaluate cosine value \[ \cos \frac{2\pi}{3} = -\frac{1}{2} \] Step 2: Substitute \[ \sin^{-1}\left(-\frac{1}{2}\right) \] Step 3: Use standard value \[ \sin\left(-\frac{\pi}{6}\right) = -\frac{1}{2} \] Step 4: Check principal range \[ -\frac{\pi}{6} \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \] —