Question

b.
Prove
tan 90-o) sec (18o-o sin (-o)/
sin (180+0) cot(360-0) cosec (90-0)=1
​

Answer 1

Answer:

sin(1800+θ)=−sinθ, cos(900+θ)=−sinθ, tan(2700−θ)=cotθ,

cot(3600−θ)=−cotθ, sin(3600−θ)=−sinθ, cos(3600+θ)=cosθ,

cosec(−θ)=−cosecθ,sin(2700+θ)=−cosθ.

∴    given exp. = (−sinθ).cosθ(−cosecθ)(−cosθ)(−sinθ).(−sinθ).cotθ(−cotθ)=1