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