Evaluate: sin-1(−√3/2) − 2sec-1(2tan π/6)
Solution:
Step 1: Evaluate sin⁻¹(−√3/2)
\[ \sin^{-1}\left(-\frac{\sqrt{3}}{2}\right) = -\frac{\pi}{3} \]
(Since range of sin-1(x) is \([- \pi/2, \pi/2]\))
Step 2: Evaluate 2tan(π/6)
\[ \tan \frac{\pi}{6} = \frac{1}{\sqrt{3}} \Rightarrow 2\tan \frac{\pi}{6} = \frac{2}{\sqrt{3}} \]
Step 3: Evaluate sec⁻¹(2/√3)
\[ \sec^{-1}\left(\frac{2}{\sqrt{3}}\right) = \frac{\pi}{6} \]
(Because sec θ = 2/√3 ⇒ cos θ = √3/2)
Step 4: Substitute
\[ -\frac{\pi}{3} – 2 \times \frac{\pi}{6} \]
\[ = -\frac{\pi}{3} – \frac{\pi}{3} = -\frac{2\pi}{3} \]
Final Answer:
Value = \[ -\frac{2\pi}{3} \]