Evaluate tan 13 π/12
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tan1213π=tan(π+12π)
Since (π+θ) is in third quadrant.By ASTC rule tanθ is positive in third quadrant.
∴tan(π+12π)=tan12π
=tan(3π−4π)
Using compound angle formula, tan(A+B)=1−tanAtanBtanA+tanB
=1−tan3πtan4πtan3π+tan4π
We know that tan4π=1 and tan3π=3 we have
=1−3×1
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