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Prove That:- tan(π/12) * tan(5π/12) * tan(7π/12) * tan(11π/12) = 1
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\[\tan(\frac{11 \pi}{12}) = \tan(\pi - \frac{ \pi}{12}) = -\tan(\frac{\pi}{12})\]
\[\tan(\frac{7 \pi}{12}) = \tan(\pi - \frac{ 5 \pi}{12}) = -\tan(\frac{5 \pi}{12})\]
yeah, i've understood dat portion...but could you plz solve the whole problem...i mean i'm nt getting it completely
Where are you not getting it?
\[LHS: [\tan(\frac{\pi}{12})\tan(\frac{5 \pi}{12})]^2\]
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yeah, then? how come the answer is 1?
\[\tan(\frac{5 \pi}{12}) = \tan(\frac{\pi}{2} - \frac{\pi}{12}) = \cot(\frac{\pi}{12})\]
ooooooooo yeah, i got it :P
\[LHS: [\tan(\frac{\pi}{12})\cot(\frac{ \pi}{12})]^2\]
Rest you know..
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thank you, very much! :)
You are welcome dear.. :)
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