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If `0 lt A lt 90^(@)` then the value of `(1)/(2) cotA[1+(secA+tanA)^(2)/(co s ecA(secA-tanA))]` |
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Answer» Correct Answer - c According to the question Put `A=45^(@)` `rArr(1)/(2)xxcot45^(@)` `[(1+(sec45^(@)-tan45^(@))^(2))/(cos sec45^(@)(sec45^ (@)-tan45^(@)))]` `rArr(1)/(2)[(1+(sqrt(2)-1)^(2))/(sqrt(2)xx(sqrt(2)-1))]` `rArr(1)/(1)[(1+2+1-2sqrt(2))/(2-sqrt(2))]` `rArr(1)/(2)[(4-2sqrt(2))/( 2-sqrt(2))]` ` rArr(1)/(2)xx2[(2-sqrt(2))/(2-sqrt(2))]` `rArr1` |
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