If sec A= 2, then find the value of `(1)/(cotA)+(cosA)/(1+sinA)`

Explanation

Given that , `secA=2=(2)/(1)`
Construct a right - angled `DeltaABC` in which
`angleABC=90^(@),AB=kandAC=2k`.
In `DeltaABC`,
From Pythagoeas theorem
`AB^(2)+BC^(2)=AC^(2)`
`rArrBC^(2)=AC^(2)-AB^(2)=(2k)^(2)-(k)^(2)=4k^(2)-k^(2)=3k^(2)`
`rArrBC=sqrt(3k)`
Now , `cot A=(AB)/(BC)=(k)/(sqrt(3k))=(1)/(sqrt(3))`
` sin A=(BC)/(AC)=(sqrt(3k))/(2k)=(sqrt(3))/(2) and cos A=(AB)/(AC)=(k)/(2k)=(1)/(2)`
`:. (1)/(cotA)+(cosA)/(1+sinA)=(1/(1))/(sqrt(3))+(1/(2))/(1+sqrt(3)/(2))=sqrt(3)+(1/(2))/((2+sqrt(3))/(2))`
`=sqrt(3)+(1xx(2-sqrt(3)))/((2+sqrt(3))(2-sqrt(3)))=sqrt(3)+(2-sqrt(3))/(4-3)`
`sqrt(3)+2-sqrt(3)=2`