`tan 20^0 tan 30^0 tan 40^0 tan 80^0=1`

Without using trigonometric tables, evaluate each of the following: (-tanthetacot(90^0-theta)+secthetacos e c(90^0-theta)+sin^2 35^0+sin^2 55^0)/(tan10^0tan20^0tan30^0tan70^0tan80^0)

Prove the following: (i) "tan" 20^(@) tan 40^(@) tan 60^(@) tan 80^(@)=3 (ii) "tan" 20^(@) tan 40^(@) tan 80^(@)="tan" 60^(@) .

Prove that : (i) (sin 70^@)/(cos 20^@)+(cosec 20^@)/(sec 70^@)-2cos 70^@cosec 20^@=0 (ii) (cos 80^@)/(sin 10^@)+cos 59^@cosec 31^@=2 (iii) (2 sin 68^@)/(cos 22^@)-(2 cot 15^@)/(5 tan 75^@)-(3tan 45^@tan 20^@tan 40^@tan 50^@tan 70^@)/(5)=1 (iv) (sin 18^@)/(cos 72^@)+sqrt(3) (tan 10^@ tan 30^@tan 40^@tan 50^@tan 80^@)=2 (v) (7cos 55^@)/(3 sin 35^@)-(4(cos 70^@cosec 20^@))/(3(tan 5^@tan 25^@tan 45^@tan 65^@tan 856^@))=1

Prove that: tan20^(@)tan30^(@)tan40^(@)tan80^(@)=1

tan20^(@)tan30^(@)tan40^(@)tan80^(@)=1

Prove that: tan10^0tan15^0tan75^0tan80^0=1 tan1^0tan2^0tan3^0tan89^0=1 cos1^0cos2^0cos3^0cos180^0=0

Evaluate: (sec theta csc(90^(0)-theta)-tan theta cot(90^(0)-theta)+sin^(2)55^(0)+sin^(2)35^(@))/(tan10^(0)tan20^(0)tan60^(@)tan70^(0)tan80^(0))

The value of tan 20^(@) tan 40^(@) tan 80^(@) is equal to