Question

Cos58/sin32+sin22/cos68 − cos90

Answer 1

Answer:

We have,

2(cos58∘sin32∘)−3–√(cos38∘cosec52∘tan15∘tan60∘tan75∘)

=2{cos(90∘−32∘)sin32∘}−3–√{cos38∘cosec(90∘−38∘)tan15∘tan60∘tan(90∘−15∘)}

=2(sin32∘sin32∘)−3–√{cos38∘sec38∘tan15∘×3√×cot15∘} [∵cos(90−θ)=sinθ,cosec(90−θ)=secθ,tan(90−θ)=cotθ]

=2−3–√{cos38∘×1cos38∘tan15∘×3√×1tan15∘}=2−3√3√=2−1=1 [secθ=1cosθ,cotθ=1tanθ]

therefore, 2(cos58∘sin32∘)−3–√(cos38∘cosec52∘tan15∘tan60∘tan75∘)=1