1.

`(tanA)/(1+secA) - (tanA)/(1-secA) = 2cosecA`

Answer» LHS =`(tanA)/(1+secA) - (tanA)/(1-secA)= (tanA(1-secA - 1 -secA))/((1+secA)(1-secA))`
`=(tanA(-2secA))/(1-sec^(2)A)=(2tan A. secA)/(sec^(2)A-1)` `[therefore (a+b)(a-b)=a^(2)-b^(2)]`
`=(2tanA. secA)/(tan^(2)A)` `[therefore sec^(2)A- tan^(2)A=1]` `[therefore sectheta=1/costheta` and `tantheta=(sintheta)/(costheta)]`
`=(2secA)/(tanA) = 2/(sinA) = 2 cosec A= RHS` `[therefore cosectheta= 1/(sintheta)]`


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