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` 

Answer 1

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