please prove cos A /1- sin A = Tan (45 +_ A/2)trignometry from SL Loney s book

We'll evaluate the right side of expression using the
tangent of a sum/difference formula:

tan(a+b) = (tan a +
tan b)/(1 - tan a*tan b)

Comparing, we'll
get:

tan(45 + A/2) = (tan 45 + tan A/2)/(1 - tan A/2) = (1
+tan A/2)/(1 - tan A/2)

Now, we'll manage the left
side,using the double angle identities:

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

The difference of two squares
will return the product:

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

1 - sin A = 1 - sin
2(A/2)

We'll use Pythagorean identity to replace
1:

(cos A/2)^2 + (sinA/2)^2 =
1

1 - sin 2(A/2) = (cos A/2)^2 + (sinA/2)^2 - 2sin(A/2)*cos
(A/2)

We'll get a perfect square at
denominator:

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

We'll re-write the left
side:

cos A/(1 - sin A) = (cos A/2-sinA/2)(cos
A/2+sinA/2)/(cos A/2+sinA/2)^2

We'll
simplify:

cos A/(1 - sin A) = (cos A/2-sinA/2)/(cos
A/2+sinA/2)

We'll factorize by cos (A/2) both numerator and
denominator:

cos A/(1 - sin A) = cos A/2*(1 -sinA/2/cos
A/2)/cos A/2*(1+sinA/2/cos A/2)

But sinA/2/cos A/2 = tan
A/2

We'll simplify:

cos A/(1 -
sin A) = (1 -tan A/22)/(1+tan A/2)

Managing
both sides, we notice that only the expression cos A/(1 - sin A) = tan(45 - A/2)
represents an identity, while the expression cos A/(1 - sin A) = tan(45 + A/2) is not an
identity.