The correct option is D 12−√32
12tan585∘+sec(−660∘)+sin405∘cot510∘−cosec (−570∘)
We know that,
tan585∘=−tan(720−585)∘=−tan135∘=1sec(−660∘)=sec(660∘)=sec(720−660)∘=sec60∘=2sin405∘=sin(360+45)∘=sin45∘=1√2cot510∘=cot(360+150)∘=cot150∘=−√3cosec (−570∘)=−cosec (570∘)=cosec (720−570)∘=cosec (150∘)=2
Now, putting the values in the expression, we get
12tan585∘+sec(−660∘)+sin405∘cot510∘−cosec (−570∘)=12+2−1√2×√3−2=12+2−√32−2=12−√32