#(sin10 sin 20 sin40 sin50)/(cos10 cos 20 cos40 cos50)# The value of that?

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Answer 1 · 2018-05-31T02:44:53

About the simplest form I found was

#sec 20^circ - 1 #

From complementary angles, #sin 50^\circ = cos 40^\circ# and vice versa, so

#{ sin 10^circ sin 20^circ sin 40^circ sin 50^circ }/{cos 10^circ cos 20^circ cos 40^circ cos 50^circ }#

# ={ sin 10^circ sin 20^circ }/ {cos 10^circ cos 20^circ} times { sin 40^circ }/{cos 50^circ } times {sin 50^circ }/ cos 40^circ #

# ={ sin 10^circ sin 20^circ }/ {cos 10^circ cos 20^circ} #

# = { sin 10^circ ( 2 \ sin 10^circ cos 10^circ) }/{cos 10^circ \cos 20^circ }#

# = {2 sin ^2 10 ^circ }/{\cos 20^circ }#

# = {1 - \cos 20^\circ}/{cos 20^\circ} #

#= sec 20^circ - 1 #