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,
\[ -\frac{\pi}{6} – \frac{\pi}{3} – \frac{\pi}{4} \]
Take LCM = 12:
\[ = -\frac{2\pi}{12} – \frac{4\pi}{12} – \frac{3\pi}{12} = -\frac{9\pi}{12} = -\frac{3\pi}{4} \]
Final Answer:
Value = \[ -\frac{3\pi}{4} \]