The correct option is C 1
cos58∘sin32∘+sin22∘cos68∘−cos38∘cosec 52∘tan18∘tan35∘tan72∘tan55∘=cos(90∘−32∘)sin32∘+sin(90∘−68∘)cos68∘−cos38∘cosec 52∘tan18∘tan35∘tan72∘tan55∘=1+1−cos38∘cosec (90∘−38∘)tan18∘tan35∘tan72∘tan55∘=2−1tan18∘tan35∘tan72∘tan55∘
Now,
tan18∘tan35∘tan72∘tan55∘=tan18∘tan35∘tan(90∘−18∘)tan(90∘−35∘)=tan18∘tan35∘cot18∘cot35∘=(tan18∘cot18∘)(tan35∘cot35∘)=1
Therefore,
cos58∘sin32∘+sin22∘cos68∘−cos38∘cosec 52∘tan18∘tan35∘tan72∘tan55∘=2−1=1