Let's analyze the given equation and see if we can transform one side to match the other.
The original equation to prove is:
[tex]\[ \frac{1 - \sin A}{\cos A} = \frac{1 - \tan (A / 2)}{1 + \tan (A / 2)} \][/tex]
First, let's work on the left-hand side (LHS):
[tex]\[ \text{LHS} = \frac{1 - \sin A}{\cos A} \][/tex]
To simplify, we need to recall the trigonometric identity for [tex]\(\sin A\)[/tex] and [tex]\(\cos A\)[/tex] using half-angle formulas. The half-angle formulas are:
[tex]\[ \sin A = 2 \sin \left(\frac{A}{2}\right) \cos \left(\frac{A}{2}\right) \][/tex]
[tex]\[ \cos A = \cos^2 \left(\frac{A}{2}\right) - \sin^2 \left(\frac{A}{2}\right) \][/tex]
Furthermore, we know that:
[tex]\[ \tan \left(\frac{A}{2}\right) = \frac{\sin \left(\frac{A}{2}\right)}{\cos \left(\frac{A}{2}\right)} = t \][/tex]
Thus,
[tex]\[ \sin \left(\frac{A}{2}\right) = t \cos \left(\frac{A}{2}\right) \][/tex]
and,
[tex]\[ \sin^2 \left(\frac{A}{2}\right) = t^2 \cos^2 \left(\frac{A}{2}\right) \][/tex]
Next, we can write [tex]\(\sin A\)[/tex] and [tex]\(\cos A\)[/tex] using [tex]\(t\)[/tex]:
[tex]\[ \sin A = 2t \cos^2 \left(\frac{A}{2}\right) \][/tex]
[tex]\[ \cos A = \cos^2 \left(\frac{A}{2}\right) - \sin^2 \left(\frac{A}{2}\right) = \cos^2 \left(\frac{A}{2}\right) (1 - t^2) \][/tex]
Now, substitute these into the LHS:
[tex]\[ \text{LHS} = \frac{1 - 2t \cos^2 \left(\frac{A}{2}\right)}{\cos^2 \left(\frac{A}{2}\right) (1 - t^2)} \][/tex]
Rewrite the numerator:
[tex]\[ 1 - 2t \cos^2 \left(\frac{A}{2}\right) = \cos^2 \left(\frac{A}{2}\right) (\frac{1}{\cos^2 \left(\frac{A}{2}\right)} - 2t) \][/tex]
Substitute back into LHS:
[tex]\[ \text{LHS} = \frac{\cos^2 \left(\frac{A}{2}\right) ( \frac{1}{\cos^2 \left(\frac{A}{2}\right)} - 2t )}{\cos^2 \left(\frac{A}{2}\right) (1 - t^2)} \][/tex]
[tex]\[ \text{LHS} = \frac{1 - 2t \cos^2 \left(\frac{A}{2}\right)}{(1 - t^2) \cos^2 \left(\frac{A}{2}\right)} \][/tex]
Canceling [tex]\(\cos^2 \left(\frac{A}{2}\right)\)[/tex]:
[tex]\[ \text{LHS} = \frac{1 - 2t}{1 - t^2} \][/tex]
Similarly, we have the right-hand side (RHS):
[tex]\[ \text{RHS} = \frac{1 - \tan \left(\frac{A}{2}\right)}{1 + \tan \left(\frac{A}{2}\right)} \][/tex]
Substituting [tex]\(\tan \left(\frac{A}{2}\right) = t\)[/tex]:
[tex]\[ \text{RHS} = \frac{1 - t}{1 + t} \][/tex]
From the above observations, we see:
[tex]\[\text{LHS} = \frac{1 - 2t}{1 - t^2} \][/tex]
[tex]\[\text{RHS} = \frac{1 - t}{1 + t} \][/tex]
In conclusion, because the simplifications eliminate the expected transformations, LHS and RHS are not equal directly. Thus,
[tex]\[ \frac{1 - \sin A}{\cos A} \neq \frac{1 - \tan (A / 2)}{1 + \tan (A / 2)} \][/tex]
Thus, the claim cannot be proven true.
