April 2026

Evaluate cot⁻¹(cot 9π/4) Problem Evaluate: \( \cot^{-1}(\cot \frac{9\pi}{4}) \) Solution First, reduce the angle using periodicity: \[ \frac{9\pi}{4} = 2\pi + \frac{\pi}{4} \] Since cotangent has period \( \pi \), we write: \[ \cot \frac{9\pi}{4} = \cot \frac{\pi}{4} \] Now, \[ \cot \frac{\pi}{4} = 1 \] Thus the expression becomes: \[ \cot^{-1}(1) \] Recall the […]

Evaluate cot⁻¹(cot 4π/3) Problem Evaluate: \( \cot^{-1}(\cot \frac{4\pi}{3}) \) Solution First, evaluate the cotangent: \[ \cot \frac{4\pi}{3} = \cot\left(\pi + \frac{\pi}{3}\right) \] Using identity: \[ \cot(\pi + \theta) = \cot \theta \] So, \[ \cot \frac{4\pi}{3} = \cot \frac{\pi}{3} = \frac{1}{\sqrt{3}} \] Thus the expression becomes: \[ \cot^{-1}\left(\frac{1}{\sqrt{3}}\right) \] Recall the principal value range of

Evaluate cot⁻¹(cot π/3) Problem Evaluate: \( \cot^{-1}(\cot \frac{\pi}{3}) \) Solution We know that: \[ \cot \frac{\pi}{3} = \frac{1}{\sqrt{3}} \] So the expression becomes: \[ \cot^{-1}\left(\frac{1}{\sqrt{3}}\right) \] Recall the principal value range of \( \cot^{-1} x \): \[ (0, \pi) \] We need an angle in this range whose cotangent is \( \frac{1}{\sqrt{3}} \). We know

Evaluate cosec⁻¹{cosec(−9π/4)} Problem Evaluate: \( \csc^{-1}(\csc(-\frac{9\pi}{4})) \) Solution First, use periodicity of sine: \[ -\frac{9\pi}{4} = -2\pi – \frac{\pi}{4} \] Since sine is periodic with period \(2\pi\): \[ \sin\left(-\frac{9\pi}{4}\right) = \sin\left(-\frac{\pi}{4}\right) \] Now, \[ \sin\left(-\frac{\pi}{4}\right) = -\frac{1}{\sqrt{2}} \] So, \[ \csc\left(-\frac{9\pi}{4}\right) = \frac{1}{\sin\left(-\frac{9\pi}{4}\right)} = -\sqrt{2} \] Thus the expression becomes: \[ \csc^{-1}(-\sqrt{2}) \] Recall the

Evaluate cosec⁻¹(cosec 13π/6) Problem Evaluate: \( \csc^{-1}(\csc \frac{13\pi}{6}) \) Solution First, reduce the angle: \[ \frac{13\pi}{6} = 2\pi + \frac{\pi}{6} \] Using periodicity: \[ \sin\left(2\pi + \theta\right) = \sin \theta \] So, \[ \sin \frac{13\pi}{6} = \sin \frac{\pi}{6} = \frac{1}{2} \] Thus, \[ \csc \frac{13\pi}{6} = \frac{1}{\sin \frac{13\pi}{6}} = 2 \] Now the expression becomes:

Evaluate cosec⁻¹(cosec 11π/6) Problem Evaluate: \( \csc^{-1}(\csc \frac{11\pi}{6}) \) Solution We know that: \[ \sin \frac{11\pi}{6} = -\frac{1}{2} \] So, \[ \csc \frac{11\pi}{6} = \frac{1}{\sin \frac{11\pi}{6}} = -2 \] Thus the expression becomes: \[ \csc^{-1}(-2) \] Recall the principal value range of \( \csc^{-1} x \): \[ \left[-\frac{\pi}{2}, 0\right) \cup \left(0, \frac{\pi}{2}\right] \] Since the

Evaluate cosec⁻¹(cosec 6π/5) Problem Evaluate: \( \csc^{-1}(\csc \frac{6\pi}{5}) \) Solution First, note that: \[ \frac{6\pi}{5} = \pi + \frac{\pi}{5} \] Using identity: \[ \sin(\pi + \theta) = -\sin \theta \] So, \[ \sin \frac{6\pi}{5} = -\sin \frac{\pi}{5} \] Thus, \[ \csc \frac{6\pi}{5} = \frac{1}{\sin \frac{6\pi}{5}} = -\frac{1}{\sin \frac{\pi}{5}} = -\csc \frac{\pi}{5} \] Now the expression

Evaluate cosec⁻¹(cosec 3π/4) Problem Evaluate: \( \csc^{-1}(\csc \frac{3\pi}{4}) \) Solution We know that: \[ \sin \frac{3\pi}{4} = \frac{1}{\sqrt{2}} \] So, \[ \csc \frac{3\pi}{4} = \frac{1}{\sin \frac{3\pi}{4}} = \sqrt{2} \] Thus the expression becomes: \[ \csc^{-1}(\sqrt{2}) \] Now recall the principal value range of \( \csc^{-1} x \): \[ \left[-\frac{\pi}{2}, 0\right) \cup \left(0, \frac{\pi}{2}\right] \] Although

Evaluate cosec⁻¹(cosec π/4) Problem Evaluate: \( \csc^{-1}(\csc \frac{\pi}{4}) \) Solution We know that: \[ \csc \frac{\pi}{4} = \frac{1}{\sin \frac{\pi}{4}} = \frac{1}{\frac{1}{\sqrt{2}}} = \sqrt{2} \] So the expression becomes: \[ \csc^{-1}(\sqrt{2}) \] Now, recall the principal value range of \( \csc^{-1} x \): \[ \left[-\frac{\pi}{2}, 0\right) \cup \left(0, \frac{\pi}{2}\right] \] We need an angle whose cosecant

Evaluate sec⁻¹(sec 25π/6) Evaluate \( \sec^{-1}(\sec \frac{25\pi}{6}) \) Step-by-Step Solution We need to evaluate: \[ \sec^{-1}\left(\sec \frac{25\pi}{6}\right) \] Step 1: Reduce the angle \[ \frac{25\pi}{6} = 4\pi + \frac{\pi}{6} \] \[ \sec\left(\frac{25\pi}{6}\right) = \sec\left(\frac{\pi}{6}\right) \] Step 2: Convert to cosine \[ \cos \frac{\pi}{6} = \frac{\sqrt{3}}{2} \Rightarrow \sec \frac{\pi}{6} = \frac{2}{\sqrt{3}} \] Step 3: Apply inverse