April 2026

Evaluate tan⁻¹(tan 1) Evaluate \( \tan^{-1}(\tan 1) \) Step-by-Step Solution We need to evaluate: \[ \tan^{-1}(\tan 1) \] Step 1: Principal value range The principal value range of \( \tan^{-1}x \) is: \[ \left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \] Step 2: Check the angle Since \( 1 \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \), it already lies in the principal range. […]

Evaluate tan⁻¹(tan 9π/4) Evaluate \( \tan^{-1}(\tan \frac{9\pi}{4}) \) Step-by-Step Solution We need to evaluate: \[ \tan^{-1}\left(\tan \frac{9\pi}{4}\right) \] Step 1: Use periodicity \[ \tan(x + \pi) = \tan x \] \[ \frac{9\pi}{4} = 2\pi + \frac{\pi}{4} \Rightarrow \tan\left(\frac{9\pi}{4}\right) = \tan\left(\frac{\pi}{4}\right) \] Step 2: Apply inverse tangent \[ \tan^{-1}\left(\tan \frac{\pi}{4}\right) \] The principal value range of

Evaluate tan⁻¹(tan 7π/6) Evaluate \( \tan^{-1}(\tan \frac{7\pi}{6}) \) Step-by-Step Solution We need to evaluate: \[ \tan^{-1}\left(\tan \frac{7\pi}{6}\right) \] Step 1: Principal value range The principal value range of \( \tan^{-1}x \) is: \[ \left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \] Step 2: Adjust the angle Since \( \frac{7\pi}{6} \in \left(\frac{\pi}{2}, \frac{3\pi}{2}\right) \), use: \[ \tan(x – \pi) = \tan

Evaluate tan⁻¹(tan 6π/7) Evaluate \( \tan^{-1}(\tan \frac{6\pi}{7}) \) Step-by-Step Solution We need to evaluate: \[ \tan^{-1}\left(\tan \frac{6\pi}{7}\right) \] Step 1: Principal value range The principal value range of \( \tan^{-1}x \) is: \[ \left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \] Step 2: Adjust the angle Since \( \frac{6\pi}{7} > \frac{\pi}{2} \), bring it into principal range using: \[ \tan(x

Evaluate tan⁻¹(tan π/3) Evaluate \( \tan^{-1}(\tan \frac{\pi}{3}) \) Step-by-Step Solution We need to evaluate: \[ \tan^{-1}\left(\tan \frac{\pi}{3}\right) \] Step 1: Principal value range The principal value range of \( \tan^{-1}x \) is: \[ \left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \] Step 2: Check the angle \[ \frac{\pi}{3} > \frac{\pi}{2} \] So it is outside the principal range. Step 3:

Evaluate cos⁻¹(cos 12) Evaluate \( \cos^{-1}(\cos 12) \) Step-by-Step Solution We need to evaluate: \[ \cos^{-1}(\cos 12) \] Step 1: Principal value range The principal value range of \( \cos^{-1}x \) is: \[ [0, \pi] \] Step 2: Reduce the angle Reduce 12 modulo \(2\pi\): \[ 12 – 2\pi \approx 12 – 6.283 = 5.717

Evaluate cos⁻¹(cos 5) Evaluate \( \cos^{-1}(\cos 5) \) Step-by-Step Solution We need to evaluate: \[ \cos^{-1}(\cos 5) \] Step 1: Principal value range The principal value range of \( \cos^{-1}x \) is: \[ [0, \pi] \] Step 2: Locate 5 Since \( 5 \in (\pi, 2\pi) \), we use identity: \[ \cos(2\pi – x) =

Evaluate cos⁻¹(cos 4) Evaluate \( \cos^{-1}(\cos 4) \) Step-by-Step Solution We need to evaluate: \[ \cos^{-1}(\cos 4) \] Step 1: Principal value range The principal value range of \( \cos^{-1}x \) is: \[ [0, \pi] \] Step 2: Locate 4 Since \( 4 \in (\pi, 2\pi) \), we use identity: \[ \cos(2\pi – x) =

Evaluate cos⁻¹(cos 3) Evaluate \( \cos^{-1}(\cos 3) \) Step-by-Step Solution We need to evaluate: \[ \cos^{-1}(\cos 3) \] Step 1: Principal value range The principal value range of \( \cos^{-1}x \) is: \[ [0, \pi] \] Step 2: Check the angle Since \( 3 \in (0, \pi) \), it already lies in the principal range.

Evaluate cos⁻¹(cos 13π/6) Evaluate \( \cos^{-1}(\cos \frac{13\pi}{6}) \) Step-by-Step Solution We need to evaluate: \[ \cos^{-1}\left(\cos \frac{13\pi}{6}\right) \] Step 1: Reduce the angle \[ \frac{13\pi}{6} = 2\pi + \frac{\pi}{6} \] \[ \cos\left(\frac{13\pi}{6}\right) = \cos\left(\frac{\pi}{6}\right) \] Step 2: Apply inverse cosine \[ \cos^{-1}\left(\cos \frac{\pi}{6}\right) \] The principal value range of \( \cos^{-1}x \) is: \[ [0,