evaluate `sin25/cos65 +cot15/tan75 + (2cos43cosec47)/(tan10tan40tan50tan80)`

Evaluate : (sin70^@)/(cos 20^@)+(cosec36^@)/(sec54^@)-(2cos43^@cosec47^@)/(tan10^@tan40^@tan50^@tan80^@ .

tan10^@tan15^@tan75^@tan80^@ is

Evaluate : (2sin68^0)/(cos22^0)-(2cot15^0)/(5tan75^0)-(3tan45^0tan20^0tan40^0tan50^0tan70^0)/5

Prove that: (2 sin 68°)/(cos 22°)-( 2 cot 15°)/( 5 tan 75°)- 3( tan 45° tan 20° tan 40 tan 50° tan 70°)/5 = 1 .

Evaluate : (2sin68^(0))/(cos22^(0))-(2cot15^(0))/(5tan75^(0))-(3tan45^(0)tan20^(@)tan40^(@)tan50^(@)tan70^(@))/(5)

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

Without using Trigonometric Tables evaluate the following :- (2sin68^@)/(cos22^@)-(2cot15^@)/(5tan75^@)-(3tan45^@tan20^@tan40^@tan70^@tan50^@)/5 .

7. (sin A+cos A)(cot A+tan A)=sec A+cosec A