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

Hint: 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.