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tan 7pi/12 + cot 5pi/12
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write \[7\pi/12=\pi/4+\pi/6~~ then\\evaluate ~~\tan(\pi/4+\pi/6)~~using~~\tan(a+b)\]
\[\tan(a+b)=\frac{ tana+tanb }{ 1-tana*tanb }\]
use this :)
and use the same procedure for cot \[\cot(a+b)=\frac{ cota*cotb-1 }{ cota+cotb }\]
so for tan \[\tan(\pi/4+\pi/6)=\frac{ \tan(\pi/4)+\tan(\pi/6) }{ 1-\tan(\pi/4)*\tan(\pi/6) }\] and fot cot\[\cot(\pi/4+\pi/6)=\frac{ \cot(\pi/4)*\cot(\pi/6)-1 }{ \cot(\pi/4)+\cot(\pi/6)}\] now just substitute the values and u'll get ur answer :)
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thanks
oops tmy mistake but yeah the process will be same \[for ~(7 \pi/12)=(\pi/4)+(\pi/3)~~and ~~for ~~5\pi/12=\pi/4+\pi/6\]
so \[\tan(\pi/4+\pi/3)+\cot(\pi/4+\pi/6)\] now jusu use those formula i wrote above
*just
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