prove the following identity- cosA/(1-tanA)+sinA/(1-cotA)=cosA+sinA

For the beginning, we'll re-write the terms from the left
side of the given expression:

cosA/(1-tanA) = cosA/(1-
sinA/cosA)

cosA/(1- sinA/cosA) =
cosA/[(cosA-sinA)/cosA]

cosA/[(cosA-sinA)/cosA] =
(cosA)^2/(cosA-sinA) (1)

sinA/(1-cotA) = sinA/(1-
cosA/sinA)

sinA/(1- cosA/sinA) = (sinA)^2/(sinA-cosA)
(2)

We'll add (1) and
(2):

(cosA)^2/(cosA-sinA) + (sinA)^2/(sinA-cosA) =
(cosA)^2/(cosA-sinA) -
(sinA)^2/(cosA-sinA)

(cosA)^2/(cosA-sinA) -
(sinA)^2/(cosA-sinA) =
[(cosA)^2-(sinA)^2]/(cosA-sinA)

We'll re-write the
difference of squares:

(cosA)^2-(sinA)^2 = (cos A -
sinA)(cos A + sinA)

The left side of the expression will
become:

(cos A - sinA)(cos A +
sinA)/(cosA-sinA)

We'll reduce the like
terms:

(cos A - sinA)(cos A +
sinA)/(cosA-sinA) = cos A + sinA q.e.d.