Question

Prove that: \[\dfrac{{\cos A}}{{1 - \sin A}} = \dfrac{{1 + \sin A}}{{\cos A}}\].

Answer 1

Hint: To solve the question, we will start with the LHS and try to get the RHS from it. We can see that in the denominator of RHS there is a cosine term and in the denominator of LHS it is the sine term. So, to get the cosine term, we will multiply the numerator and denominator of LHS by \[1 + \sin A\]. Then, with the use of trigonometry we will solve further to get the RHS.

Complete step-by-step solution:
LHS: \[\dfrac{{\cos A}}{{1 - \sin A}}\]
Now in order to get the cosine term in the denominator we will multiply both the numerator and denominator by \[1 + \sin A\]. So, we get;
\[ = \dfrac{{\cos A}}{{1 - \sin A}} \times \dfrac{{1 + \sin A}}{{1 + \sin A}}\]
Now we can see that the denominator is in the form \[\left( {a + b} \right)\left( {a - b} \right)\] and we know that:
\[\left( {a + b} \right)\left( {a - b} \right) = {a^2} - {b^2}\]
Therefore,
\[\left( {1 - \sin A} \right)\left( {1 + \sin A} \right) = 1 - {\sin ^2}A\]
So, we get;
\[ = \dfrac{{\cos A\left( {1 + \sin A} \right)}}{{1 - {{\sin }^2}A}}\]
Now we know that, \[{\sin ^2}A + {\cos ^2}A = 1\]
\[\therefore 1 - {\sin ^2}A = {\cos ^2}A\]
Now we will use this relation in the above fraction. So, we get;
\[ = \dfrac{{\cos A\left( {1 + \sin A} \right)}}{{{{\cos }^2}A}}\]
Now we will cancel the cosine term from both the numerator and denominator. So, we get,
\[ = \dfrac{{\left( {1 + \sin A} \right)}}{{\cos A}}\]
And this is equal to the RHS.

Note: In this question one can also think of starting with the RHS and try to get the LHS in the process. But either that will be a very lengthy approach or one can not arrive at the result.
We can also solve this by another simple method that is by dividing the LHS by \[\cos A\] and applying the formula that: \[{\sec ^2}A - {\tan ^2}A = 1\].
LHS: \[\dfrac{{\cos A}}{{1 - \sin A}}\]
Dividing the numerator and denominator by \[\cos A\].
\[ = \dfrac{{\left( {\dfrac{{\cos A}}{{\cos A}}} \right)}}{{\dfrac{{1 - \sin A}}{{\cos A}}}}\]
Simplifying we get;
\[ = \dfrac{1}{{\left( {\dfrac{1}{{\cos A}} - \dfrac{{\sin A}}{{\cos A}}} \right)}}\]
Now we will write \[\dfrac{1}{{\cos A}} = \sec A\] and \[\dfrac{{\sin A}}{{\cos A}} = \tan A\]. So, we get;
\[ = \dfrac{1}{{\sec A - \tan A}}\]
Now we will multiply and divide by \[\sec A + \tan A\].
\[ = \dfrac{{\sec A + \tan A}}{{\left( {\sec A + \tan A} \right)\left( {\sec A - \tan A} \right)}}\]
On simplification we get;
\[ = \dfrac{{\sec A + \tan A}}{{{{\sec }^2}A - {{\tan }^2}A}}\]
Now we will use \[{\sec ^2}A - {\tan ^2}A = 1\]. So, we get,
\[ = \sec A + \tan A\]
Writing in the form of sine and cosine. We get,
\[ = \dfrac{1}{{\cos A}} + \dfrac{{\sin A}}{{\cos A}}\]
Taking the LCM and solving we get;
\[ = \dfrac{{1 + \sin A}}{{\cos A}}\]
And this is the RHS.