Correct Answer - a
`tan15^(@)cot75^(@)+tan75^(@)cot15^(@)`
`=tan15^(@)cot (90^(@)-15^(@))+tan (90^(@)-15^(@)).cot5^(@)`
`=tan^(2)15^(@)+cot^(2)15^(@)=tan^(2)15 ^(@)+cot^(2)15^(@)`..........(i)
Formula
`cot(90^(@)-theta)=tan theta`
`tan(90^(@)-theta)=cot theta`
Put value of `tan 15^(@)`
`[cot 15^(@)=(1)/(tan15^(@)) cot15^(@)=(1)/((2-sqrt(3)))rArr(1)/((2-sqrt(3)))xx ((2+sqrt(3)))/((2+sqrt(3)))]`
`cot1 8^(@)=2+sqrt(3)`
Now put value in eq (i)
`tan^(2) 15^(@)cot^(2)15^(@)`
`= (2-sqrt(3))^(2)+(2+sqrt(3))^(2)`
`=4+3-4sqrt(3)+4+3+4sqrt(3)`
=14
