1.

`sec70^(@)sin20^(@)+cos20^(@)cosec70^(@)=2`

Answer» `sec70^(@)sin(90^(@)-70^(@))+cos20^(@)cosec(90^(@)-20^(@))`
`=sec70^(@)cos70^(@)+cos20^(@)sec20^(@)`
`=(1)/(cos70^(@)).cos70^(@)+cos20^(@).(1)/(cos20^(@))`
=1+1=2.
Hence `sec70^(@)sin20^(@)+cos20^(@)cosec70^(@)=2`.


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