Question

Prove that $\dfrac{1}{{\sec {\text{A}} - \tan {\text{A}}}} - \dfrac{1}{{\cos {\text{A}}}} = \dfrac{1}{{\cos {\text{A}}}} - \dfrac{1}{{\sec {\text{A + tanA}}}}$

Answer 1

Here in the above given problem we are given $\dfrac{1}{{\sec {\text{A}} - \tan {\text{A}}}} - \dfrac{1}{{\cos {\text{A}}}} = \dfrac{1}{{\cos {\text{A}}}} - \dfrac{1}{{\sec {\text{A + tanA}}}}$ and by the use of some basic formula and identity like$[(a + b)(a - b) = {a^2} - {b^2}]$, and ${\sec ^2}{\text{A}} - {\tan ^2}{\text{A}} = 1$we will prove that Left Hand Side is equal to Right Hand Side. Lastly by cancelling the common terms we get desired results.

Formula used:
In the above question we have used the rationalization and elimination method, and rules that $[(a + b)(a - b) = {a^2} - {b^2}]$and ${\sec ^2}{\text{A}} - {\tan ^2}{\text{A}} = 1$to get the desired result.

Complete step-by-step answer:
As asked we need to prove $\dfrac{1}{{\sec {\text{A}} - \tan {\text{A}}}} - \dfrac{1}{{\cos {\text{A}}}} = \dfrac{1}{{\cos {\text{A}}}} - \dfrac{1}{{\sec {\text{A + tanA}}}}$
So firstly we will begin with Left Hand side where $LHS = \dfrac{1}{{\sec {\text{A}} - \tan {\text{A}}}} - \dfrac{1}{{\cos {\text{A}}}}$
Now here through rationalization and by multiplying and dividing the numerator and denominator by $\sec {\text{A + tan A}}$and further simplification we obtain equation as

$$\Rightarrow \dfrac{{\sec {\text{A + tanA}}}}{{(\sec {\text{A + tanA)(secA - tanA)}}}} - \dfrac{1}{{\cos {\text{A}}}} \\\ \Rightarrow \sec {\text{A + tanA - secA}} \\\$$

As we know that ${\sec ^2}{\text{A}} - {\tan ^2}{\text{A}} = 1$
So lastly we get $\tan {\text{A}}$
Now by adding and subtracting $\sec {\text{A}}$we get

$$\Rightarrow \sec {\text{A + tanA - secA}} \\\ \Rightarrow = \dfrac{1}{{\cos {\text{A}}}} - (\sec {\text{A - tanA)}} \\\$$

Later on by multiplying and dividing $(\sec {\text{A - tanA)}}$by $(\sec {\text{A + tanA)}}$and further simplification by cancelling out the lie terms we get the following equation

$$\Rightarrow \dfrac{1}{{\cos {\text{A}}}} - \dfrac{{({{\sec }^2}{\text{A - ta}}{{\text{n}}^2}{\text{A)}}}}{{(\sec {\text{A + tanA)}}}} \\\ \Rightarrow \dfrac{1}{{\cos {\text{A}}}} - \dfrac{1}{{\sec {\text{A + tanA}}}} \\\$$

As we can see that on LHS we have $\dfrac{1}{{\cos {\text{A}}}} - \dfrac{1}{{\sec {\text{A + tanA}}}}$ and on RHS also we have$\dfrac{1}{{\cos {\text{A}}}} - \dfrac{1}{{\sec {\text{A + tanA}}}}$which means left hand side is equal to Right Hand Side.
Hence the above given question is proved.

Note: Keep in mind that when we solve questions like this we use the most important identity that is $[(a + b)(a - b) = {a^2} - {b^2}]$and the trigonometric identity ${\sec ^2}{\text{A}} - {\tan ^2}{\text{A}} = 1$.Remember that while simplifying such questions we need to very careful because if we get missed in the initial stage we will get wrong solution. We only need to remember the basic formula of trigonometric for understanding such questions and solving it with ease.