# cosA /1-tanA +sinA / 1-cotA =sinA+cosA

the second part would be:

sin^2A / (sinA - cosA)
cos2A / (cos A - sin A) + sin2A / (sinA - cosA)
= cos2A / (cos A - sin A) - sin2A / (cos A - sin A)
= (cos2A - sin2 A)/(cos A - sin A)
= (cos A - sin A)(cos A + sin A)/ (cos A - sin A)
= cos A + sin A

• -12

First off, work on the complicated side, which is the LHS.

Usually if we see tan A and cot A in an expression together with sin A and cos A, we change tan A to sin A/cos A and cot A to cos A/ sin A before further simplifying the expression.

• -15

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

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

then,
(cos^2A - sin^2A / cosA) + (sin^2A - cos^2A / sinA) = 0
which is also not correct

Your orginal question is to prove
cosA / (1 - tanA) + sinA / (1 - cotA) = Sin A + cosA

Let's focus on cosA / (1 - tanA) first. After changing tan A to sin A/cos A, what do you get ? (Hint: you need to simply the denominator first)

• -8

1-tan A
= 1- (sin A / cos A)
= (cos A - sin A) / cos A

therefore,
cos A / (1- tan A)
= cos A / [(cos A - sin A) / cos A]
= cos2A / (cos A - sin A)

• -12

L.H.S = (1 + cos A / sin A + sin A / cos A ) (sin A -cos A)
=>((sin A * cos A + cos A * cos A + sin A * sin A) / sin A * cos A) (sin A -cos A)
=> (sin (cube) A - cos (cube) A ) / sin A * cos A
=> sin (square) A / cos A - cos (square) A / sin A
= RHS

• -11
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