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Question
Prove that:
`(2 "sin" 68^circ)/(cos 10^circ )- (2 cot 15^circ)/(5 tan 75^circ) = ((3 tan 45^circ t an 20^circ tan 40^circ tan 50^circ tan 70^circ)) /5= 1`
Solution
LHS = `(2 "sin" 68^circ)/(cos 10^circ )- (2 cot 15^circ)/(5 tan 75^circ) = ((3 tan 45^circ tan 20^circ tan 40^circ tan 50^circ tan 70^circ)) /5`
`=(2 sin 68^circ)/sin(90^circ - 22^circ) - (2 cot 15^circ)/(5 cot (90^circ - 75^circ)) - 3xx1xxcot(90^circ-20^circ)xxcot(90^circ-40^circ)xxtan 50^circxxtan 70^circ`
`= 2 - 2/5 = (3xx1/(tan 70^circ)xx1/(tan 50)^circxxtan 70^circ )/5`
`= 2 - 2/5 = 3/5`
`=( 10 - 2 - 3)/5`
`= 5/5`
= 1
= RHS
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