Question

How do you prove $ \dfrac{{\sec a - 1}}{{{{\sin }^2}a}} = \dfrac{{{{\sec }^2}a}}{{1 + \sec a}} $ ?

Answer 1

Hint: Multiply $ {\sec ^2}a $ to both numerator and denominator. To solve trigonometric problems like this, we sometimes have to do changes which modify the given LHS or RHS in such a way that, if we continue solving the new LHS or RHS then we can easily get the other side of the equation . That’s why we will multiply $ \sec a + 1 $ to both the numerator and denominator of LHS. After that we will open the brackets in numerator and multiply the terms. After that we will replace $ {\sec ^2}a - 1 $ by $ {\tan ^2} $ . Post this step, we will replace $ {\tan ^2} a $ by $ \dfrac{{{{\sin }^2}a}}{{{{\cos }^2}a}} $ . Cancelling out the common terms, we will then write $ \dfrac{1}{{{{\cos }^2}a}} $ as $ {\sec ^2}a $ which will be our RHS.

Complete step-by-step solution:
The given question we have is $ \dfrac{{\sec a - 1}}{{{{\sin }^2}a}} = \dfrac{{{{\sec }^2}a}}{{1 + \sec a}} $
Before moving forward, we will need to remember a few trigonometric formulas which we will use in this question. Those are:
 $
  {\sec ^2}a - 1 = {\tan ^2}a \\
  \tan a = \dfrac{{\sin a}}{{\cos a}} \\
  \dfrac{1}{{\cos a}} = \sec a \\
 $
In the given question, LHS is $ \dfrac{{\sec a - 1}}{{{{\sin }^2}a}} $ and RHS is $ \dfrac{{{{\sec }^2}}}{{1 + \sec a}} $
In order to prove our question, all we have to do is start with either LHS or RHS and start solving it to get the opposite side.
So starting off with LHS, we will multiply and divide $ \sec a + 1 $ to both numerator and denominator.
 $
\Rightarrow \dfrac{{\sec a - 1}}{{{{\sin }^2}a}} = \dfrac{{\left( {\sec a - 1} \right)\left( {\sec a + 1} \right)}}{{\left( {{{\sin }^2}a} \right)\left( {\sec a + 1} \right)}} \\
\Rightarrow \dfrac{{\sec a - 1}}{{{{\sin }^2}a}} = \dfrac{{{{\sec }^2}a - 1}}{{{{\sin }^2}a\left( {\sec a + 1} \right)}} \\
 $
Before moving forward, we will need to use another trigonometric formula to replace the numerator. Which is
 $ \Rightarrow {\sec ^{^2}}a - 1 = {\tan ^2}a $
So our new LHS becomes
\[\Rightarrow \dfrac{{{{\sec }^2}a - 1}}{{\left( {{{\sin }^2}a} \right)\left( {\sec a + 1} \right)}} = \dfrac{{{{\tan }^2}a}}{{{{\sin }^2}a(\sec a + 1)}}\]
Again, we know that
\[
 \Rightarrow \tan a = \dfrac{{\sin a}}{{\cos a}} \\
  \therefore {\tan ^2}a = \dfrac{{{{\sin }^2}a}}{{{{\cos }^2}a}} \\
\]
Replacing the vale, our new LHS will be
\[
  \Rightarrow \dfrac{{{{\tan }^2}a}}{{{{\sin }^2}a(\sec a + 1)}} = \dfrac{{\dfrac{{{{\sin }^2}a}}{{{{\cos }^2}a}}}}{{{{\sin }^2}a(\sec a + 1)}} \\
\Rightarrow \dfrac{{{{\tan }^2}a}}{{{{\sin }^2}a(\sec a + 1)}} = \dfrac{{{{\sin }^2}a}}{{{{\cos }^2}a}} \times \dfrac{1}{{{{\sin }^2}a(\sec a + 1)}} \\
 \Rightarrow \dfrac{{{{\tan }^2}a}}{{{{\sin }^2}a(\sec a + 1)}} = \dfrac{1}{{{{\cos }^2}a(\sec a + 1)}} \\
 \]
Replacing \[{\cos ^2}a\] by \[\dfrac{1}{{{{\sec }^2}a}}\]. New LHS will be
\[\dfrac{1}{{{{\cos }^2}a(\sec a + 1)}} = \dfrac{{{{\sec }^2}a}}{{\sec a + 1}} = RHS\]

Hence, when we solve the LHS of the equation, we end up with a value which was our RHS. This thing proves that RHS=LHS and hence the question given to us is finally proved.

Note: Properties of trigonometry is one of the most important things in the chapter. You can’t solve any question in this chapter without knowing properties. It is a must if you want to ace every trigonometry question. Therefore, try to remember them all and use them strategically to solve all of your problems.