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

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

We know that:

tanA = sinA/cosA

cotA = cosA/sinA

Now substitute in L.H.S:

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

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

= (cos^2 A - sin^2 A)/ (cosA-sinA)

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

= cosA + sinA = R.H.S

To prove cosA/(1-tanA)+sinA/(1-cotA) = cosA+sinA.

Proof:

CotA = cosA/sinA and tanA = sinA. Make these substitution in LHS:

sinA/(1- sinA/cosA) +CosA/(1-cosA/sinA)

cosA*sinA/(cosA-sinA)+cosAsinA/(sinA-cosA)

cosAsinA/(cosA-sinA) - cosAsinA/(cosA-sinA)

= 0.

So the LHS = 0. But the RHS is an expression sinA+cosA. This cannot be an identity, but could only be an equation for sinA+cosA = 0

Or sinA* 1/sqrt2 +cosA*1/sqrt2 = 0

sinA*cospi/4+cosA*sinpi/4 = 0

sin (A+pi/4) =0

A+pi/4 = npi

A = npi-pi/4 = (4n-1)pi/4, n =0,1,2.....

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.**

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

L.H.S=cosA/(1-tanA)+sinA/(1-cotA)

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

=(sinA+cosA)cos^2 A+(sinA+cosA)sin^2 A because (1-tan^2 A=sec^2 A) & (1-cot^2 A=csc^2 A)

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

=sinA+cosA (because cos^2 A+sin^2 A=1)

=R.H.S