Question

cot 91* cot92* cot93* cot94*...............*cot 178* cot 179
What is the answer? Plz describe with proper formulae

Answer 1

write cot as cos/sin ... and , write all those terms as (90+x) on numerators and (180-x) on denominators and see what you get.

Answer 2

\[\frac{ \cos(90+1).\cos(90+2).......\cos(90+88). \cos(90+89 }{ \sin(180-89). \sin(180-88).....\sin(180-2).\sin(180-1) }\] whats aftr this?? @experimentX

Answer 3

yep ... what is value of \( cos(90+x) \) in terms of sin
and what is value of \( \sin(180-x) \) in term of sin ??

Answer 4

dont know...!! @experimentX

Answer 5

cos(90+x) = -sin(x)
sin(190-x) = sin(x)

Answer 6

you know cos(A+B) and sin(A+B) formula?

Answer 7

yes @experimentX

Answer 8

yep ... use it to show
cos(90+x) = -sin(x)
sin(190-x) = sin(x)
and use it on your problem.

Answer 9

is the answer -1 @experimentX

Answer 10

yep

Answer 11

thnQ @experimentX

Answer 12

yw