If `0 lt A lt 90^(@)` then the value of `(1)/(2) cotA[1+(secA+tanA)^(2)/(co s ecA(secA-tanA))]`

Options

A. 0
B. 2
C. 1
D. `(1)/(2)`

Answer

c

Explanation

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`