Question

If $\tan A/2 = 3/2$, then $(1 + \cos A)/(1 - \cos A) =$
$1$) $- 5$
$2$) $5$
$3$) $9/4$
$4$) $4/9$

Answer 1

In the question the value of tangent of a half angle ($\tan A$) is given and need to find the value of $\dfrac{{1 + \cos A}}{{1 - \cos A}}$. Since a tangent of a half angle is given, let's convert the cosine also in the form of half angle for further simplification ($\cos A$).
For that we have the formulas for sine, cosine, and tangent of half an angle. By using the formula: ${\sin ^2}A + {\cos ^2}A = 1$, formulas for sine, cosine, and tangent of half an angle will be evolve, that is, by converting in terms of half angle the formula will become as ${\sin ^2}\left( {\dfrac{A}{2}} \right) + {\cos ^2}\left( {\dfrac{A}{2}} \right) = 1$.

Formula used:
$\cos A = {\cos ^2}\left( {\dfrac{A}{2}} \right) - {\sin ^2}\left( {\dfrac{A}{2}} \right)$
By substituting ${\cos ^2}\left( {\dfrac{A}{2}} \right) - 1$ in terms of ${\sin ^2}\left( {\dfrac{A}{2}} \right)$,
$\cos A = 2{\cos ^2}\left( {\dfrac{A}{2}} \right) - 1$.
Similarly substituting $1 - {\sin ^2}\left( {\dfrac{A}{2}} \right)$ in terms of ${\cos ^2}\left( {\dfrac{A}{2}} \right)$,
$\cos A = 1 - 2{\sin ^2}\left( {\dfrac{A}{2}} \right)$.

Complete answer:
Given that the half angle of a tangent is: $\tan \dfrac{A}{2} = \dfrac{3}{2}$.
Here, $\dfrac{{1 + \cos A}}{{1 - \cos A}}$ in terms of $\cos A$, let substitute $2{\cos ^2}\left( {\dfrac{A}{2}} \right) - 1$ in the numerator and $1 - 2{\sin ^2}\left( {\dfrac{A}{2}} \right)$ in the denominator,
$\dfrac{{1 + \cos A}}{{1 - \cos A}} = \dfrac{{1 + 2{{\cos }^2}\left( {\dfrac{A}{2}} \right) - 1}}{{1 - \left( {1 - 2{{\sin }^2}\left( {\dfrac{A}{2}} \right)} \right)}}$
$+ 1$ and $- 1$ in the numerator will get cancel and by multiplying $-$ inside the brackets of values in the denominator we will get,
$= \dfrac{{2{{\cos }^2}\left( {\dfrac{A}{2}} \right)}}{{1 - 1 + 2{{\sin }^2}\left( {\dfrac{A}{2}} \right)}}$
$+ 1$ and $- 1$ in the denominator will get cancel,
$= \dfrac{{2{{\cos }^2}\left( {\dfrac{A}{2}} \right)}}{{2{{\sin }^2}\left( {\dfrac{A}{2}} \right)}}$
$2$ in both the numerator and denominator will get cancel,
$= \dfrac{{{{\cos }^2}\left( {\dfrac{A}{2}} \right)}}{{{{\sin }^2}\left( {\dfrac{A}{2}} \right)}}$
We know that $\dfrac{{\cos A}}{{\sin A}} = \cot A$, then $\dfrac{{{{\cos }^2}\left( {\dfrac{A}{2}} \right)}}{{{{\sin }^2}\left( {\dfrac{A}{2}} \right)}} = {\cot ^2}\left( {\dfrac{A}{2}} \right)$, by substituting this,
$= {\cot ^2}\left( {\dfrac{A}{2}} \right)$
We know $\cot A = \dfrac{1}{{\tan A}}$, therefore ${\cot ^2}\left( {\dfrac{A}{2}} \right) = \dfrac{1}{{{{\tan }^2}\left( {\dfrac{A}{2}} \right)}}$, by applying,
$= \dfrac{1}{{{{\tan }^2}\left( {\dfrac{A}{2}} \right)}}$
In the question we have the value of $\tan \left( {\dfrac{A}{2}} \right)$, substituting the value $\dfrac{3}{2}$ for ${\tan ^2}\left( {\dfrac{A}{2}} \right)$, we will get
$= \dfrac{1}{{{{\left( {\dfrac{3}{2}} \right)}^2}}}$
$= \dfrac{1}{{\dfrac{9}{4}}}$
By taking the reciprocal of denominator,
$= 1 \times \dfrac{4}{9}$
$= \dfrac{4}{9}$.
Thus, $(1 + \cos A)/(1 - \cos A) = 4/9$.
Hence, option (4) $4/9$ is correct.

Note:
One who follows this simplification may get doubt why should we particularly substitute $2{\cos ^2}\left( {\dfrac{A}{2}} \right) - 1$ in the numerator and $1 - 2{\sin ^2}\left( {\dfrac{A}{2}} \right)$ in the denominator, since both are the equals of $\cos A$. It's not necessary to substitute like this, we can also substitute $1 - 2{\sin ^2}\left( {\dfrac{A}{2}} \right)$ in the numerator and $2{\cos ^2}\left( {\dfrac{A}{2}} \right) - 1$ in the denominator. However the final result will be the same only in both ways.