Question

What is the value of $\dfrac{{1 - \cos A}}{{1 + \cos A}}$ ?

Answer 1

In order to solve the given question, first of all we will use the double angle trigonometric identities that are: $\cos 2x = 2{\cos ^2}x - 1$ and $\cos 2x = 1 - 2{\sin ^2}x$ .Here according to the problem, we will replace $2x$ by $A$ and proceed further through the problem. After that we will use the identity $\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}$ and then we will finally use the identity $1 + {\tan ^2}\theta = {\sec ^2}\theta$ and make the necessary calculations to get the required answer.

Complete step by step answer:
According to the problem, we are asked to find the value of $\dfrac{{1 - \cos A}}{{1 + \cos A}}$. Let it be as an equation $\left( i \right)$. Therefore, we have
$\dfrac{{1 - \cos A}}{{1 + \cos A}}{\text{ }} - - - \left( i \right)$
Now we know that according to the trigonometric double angle identities:
$\cos 2x = 2{\cos ^2}x - 1$
which can also be written as
$\Rightarrow 1 + \cos 2x = 2{\cos ^2}x{\text{ }} - - - \left( {ii} \right)$

And
$\cos 2x = 1 - 2{\sin ^2}x$
which can also be written as
$\Rightarrow 1 - \cos 2x = 2{\sin ^2}x{\text{ }} - - - \left( {iii} \right)$
Now the given problem is in the form of angle $A$ so we will replace $2x$ by $A$
$\Rightarrow x = \dfrac{A}{2}$
Therefore, from the equation $\left( {ii} \right)$ we have
$\Rightarrow 1 + \cos A = 2{\cos ^2}\dfrac{A}{2}{\text{ }} - - - \left( a \right)$
And from the equation $\left( {iii} \right)$ we have
$\Rightarrow 1 - \cos A = 2{\sin ^2}\dfrac{A}{2}{\text{ }} - - - \left( b \right)$

Now on substituting the values from the equation $\left( a \right)$ and $\left( b \right)$ in the equation $\left( i \right)$ we get
$\Rightarrow \dfrac{{1 - \cos A}}{{1 + \cos A}} = \dfrac{{2{{\sin }^2}\dfrac{A}{2}}}{{2{{\cos }^2}\dfrac{A}{2}}}$
On cancelling $2$ from both numerator and denominator we get
$\Rightarrow \dfrac{{1 - \cos A}}{{1 + \cos A}} = \dfrac{{{{\sin }^2}\dfrac{A}{2}}}{{{{\cos }^2}\dfrac{A}{2}}}$
$\Rightarrow \dfrac{{1 - \cos A}}{{1 + \cos A}} = {\left( {\dfrac{{\sin \dfrac{A}{2}}}{{\cos \dfrac{A}{2}}}} \right)^2}$

Now we know that the trigonometric identity of $\tan \theta$ in the form of $\sin \theta$ and $\cos \theta$ is given as:
$\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}$
Therefore, using it in the above equation, we have
$\Rightarrow \dfrac{{1 - \cos A}}{{1 + \cos A}} = {\left( {\tan \dfrac{A}{2}} \right)^2}$
Now we know that
$1 + {\tan ^2}\theta = {\sec ^2}\theta$
Therefore, we get
$\therefore \dfrac{{1 - \cos A}}{{1 + \cos A}} = {\sec ^2}\dfrac{A}{2} - 1$
Which is the required result.

Hence, the value of $\dfrac{{1 - \cos A}}{{1 + \cos A}}$ is ${\sec ^2}\dfrac{A}{2} - 1$.

Note: Whenever we get this type of problem, we first try to find the values of the independent variable at which the given function is not valid which makes a huge difference in the required solution. Also, the main key formula to remember is the double angle formula of trigonometric function. Also, in this type of problem, transforming one function into another function is the key concept. So, do it correctly and avoid calculation mistakes.