Question

Without using trigonometric tables , evaluate : `((tan20^(@))/(cosec70^(@)))^(2)+((cot20^(@))/(sec70^(@)))^2+2tan 15^(@)tan37^(@)tan53^(@)tan60^(@)tan75^(@)`

Answer 1

`((tan20^(@))/("cosec"70^(@)))^(2)+((cot20^(@))/(sec70^(@)))=+2tan 15^(@)tan37^(@)tan53^(@)60^(@)tan75^(@)`
`{(tan20^(@))/("cosec"(90^(@)-20^(@)))}^(2)+{(cot20^(@))/(sec(90^(@)-20^(@)))}`
`+2tan15^(@)tan37^(@)tan(90^(@)-37^(@))*(sqrt(3))tan(90^(@)-15^(@))`
`((tan20^(@))/(sec20^(@)))^(2)+((cot20^(@))/("cosec "20^(@)))^(2)+2tan15^(@)tan37^(@)cot37^(@)*(sqrt(3))cot15^(@)`
`=((sin20^(@)//cos20^(@))/(1//cos20^(@)))^(2)+((cos20^(@)//sin20^(@))/(1//sin20^(@)))^(2)+2sqrt(3)tan15^(@)tan37^(@)*(1)/(tan37^(@))*(1)/(tan15^(@))`
`sin^(2)20^(@)+cos^(2)20^(@)+2sqrt(3)=1+2sqrt(3)`