Question

Prove that tan62° −cot48° = cot28° − tan42°

Answer 1

Answer:

using trigonometric identities tan(90°-A)=cotA and cot(90°–A)=tanA,we get.

tan62°– cot48°=tan(90°-28°) -cot(90°-42°)

tan62°– cot48°=cot28° –tan42°,Proved

Answer 2

taking LHS

Tan62 -cot48

since......tanA=cot(90-A)

cot(90-62)-cot48

cot28-cot48

also.....cotA=tan(90-A)

cot28-tan(90-48)

cot28-tan42

bye