Educational

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:

Principal Value of tan⁻¹{2sin(4cos⁻¹(√3/2))} Evaluate: tan-1{2sin(4cos-1(√3/2))} Solution: Given: \[ y = \tan^{-1}\left(2\sin\left(4\cos^{-1}\frac{\sqrt{3}}{2}\right)\right) \] Step 1: Evaluate cos-1(√3/2) \[ \cos^{-1}\left(\frac{\sqrt{3}}{2}\right) = \frac{\pi}{6} \] Step 2: Substitute \[ y = \tan^{-1}\left(2\sin\left(4 \cdot \frac{\pi}{6}\right)\right) = \tan^{-1}\left(2\sin\left(\frac{2\pi}{3}\right)\right) \] Step 3: Evaluate sin(2π/3) \[ \sin\frac{2\pi}{3} = \frac{\sqrt{3}}{2} \] So, \[ 2 \times \frac{\sqrt{3}}{2} = \sqrt{3} \] Step 4: Final evaluation

Principal Value of tan⁻¹(−1) + cos⁻¹(−1/√2) Evaluate: tan-1(−1) + cos-1(−1/√2) Solution: Using principal values: \[ \tan^{-1}(-1) = -\frac{\pi}{4} \] (Range of tan-1(x) is \((- \pi/2, \pi/2)\)) \[ \cos^{-1}\left(-\frac{1}{\sqrt{2}}\right) = \frac{3\pi}{4} \] (Range of cos-1(x) is \([0, \pi]\)) Now, \[ \tan^{-1}(-1) + \cos^{-1}\left(-\frac{1}{\sqrt{2}}\right) = -\frac{\pi}{4} + \frac{3\pi}{4} \] \[ = \frac{2\pi}{4} = \frac{\pi}{2} \] Final Answer:

Principal Value of tan⁻¹(2cos 2π/3) Find the Principal Value of tan-1(2cos 2π/3) Solution: Given: \[ y = \tan^{-1}\left(2\cos \frac{2\pi}{3}\right) \] Step 1: Evaluate cos(2π/3) \[ \cos \frac{2\pi}{3} = -\frac{1}{2} \] So, \[ 2\cos \frac{2\pi}{3} = 2 \times \left(-\frac{1}{2}\right) = -1 \] Step 2: Use inverse tangent \[ y = \tan^{-1}(-1) \] We know: \[ \tan^{-1}(-1)