Question

How do you prove $\dfrac{1}{{\sec x - \tan x}} - \dfrac{1}{{\sec x + \tan x}} = 2\tan x$?

Answer 1

Hint: In order to prove the given expression, we will first consider the complex side of the equation, simplify it using the different identities and formulae. Convert the $\sec x$ and $\tan x$ into expressions with $\sin x$ and $\cos x$, and solve. Through successful application of various identities and formulae, we simplify the expression until we get the same expression as on the other side of the equation.

Complete step-by-step solution:
We need to prove: $\dfrac{1}{{\sec x - \tan x}} - \dfrac{1}{{\sec x + \tan x}} = 2\tan x$
Proof:
First, we consider the complex side of the equation, i.e., the LHS:
$\dfrac{1}{{\sec x - \tan x}} - \dfrac{1}{{\sec x + \tan x}}$ ………………………..(1)
Again, we know that:
$\sec x = \dfrac{1}{{\cos x}}$ ……………………….(2)
And $\tan x = \dfrac{{\sin x}}{{\cos x}}$ …………………….(3)
Thus, substituting the values of $\sec x$ and $\tan x$ from equations (2) and (3) in expression (1), we get:
$\Rightarrow \dfrac{1}{{\left( {\dfrac{1}{{\cos x}} - \dfrac{{\sin x}}{{\cos x}}} \right)}} - \dfrac{1}{{\left( {\dfrac{1}{{\cos x}} + \dfrac{{\sin x}}{{\cos x}}} \right)}}$
Then, on simplifying the denominators, we get:
$\Rightarrow \dfrac{1}{{\left( {\dfrac{{1 - \sin x}}{{\cos x}}} \right)}} - \dfrac{1}{{\left( {\dfrac{{1 + \sin x}}{{\cos x}}} \right)}}$
$\Rightarrow \dfrac{{\cos x}}{{1 - \sin x}} - \dfrac{{\cos x}}{{1 + \sin x}}$
Now, taking $\cos x$ as common from both the above terms, we get:
$\Rightarrow \cos x\left( {\dfrac{1}{{1 - \sin x}} - \dfrac{1}{{1 + \sin x}}} \right)$
Then, we rewrite the terms within brackets as a single term:
$\Rightarrow \cos x\left[ {\dfrac{{1 + \sin x - \left( {1 - \sin x} \right)}}{{\left( {1 - \sin x} \right)\left( {1 + \sin x} \right)}}} \right]$ …………………..(4)
We know that, $\left( {a - b} \right)\left( {a + b} \right) = {a^2} - {b^2}$
Thus, the denominator of the expression (4) can be rewritten as:
$\Rightarrow \left( {1 - \sin x} \right)\left( {1 + \sin x} \right) = 1 - {\sin ^2}x$ ……………………….(5)
Therefore, after substituting the value of $\left( {1 - \sin x} \right)\left( {1 + \sin x} \right)$ from equation (5) in expression (4), and also simplifying the numerator, we get:
$\Rightarrow \cos x\left( {\dfrac{{1 + \sin x - 1 + \sin x}}{{1 - {{\sin }^2}x}}} \right)$
Now, in the numerator of the above expression, both the ‘1’s cancel out and also there are two $\sin \left( x \right)$ terms. Thus, we have:
$ \Rightarrow \cos x\left( {\dfrac{{2\sin x}}{{1 - {{\sin }^2}x}}} \right)$ ………………………(6)
Again, we have a trigonometric identity as: ${\sin ^2}x + {\cos ^2}x = 1$ ……………………..(7)
On rearranging equation (7), we get:
$\Rightarrow 1 - {\sin ^2}x = {\cos ^2}x$ ……………………..(8)
Now, substituting the value of $\left( {1 - {{\sin }^2}x} \right)$ from equation (8) in expression (7), we get:
$\Rightarrow \cos x\left( {\dfrac{{2\sin x}}{{{{\cos }^2}x}}} \right)$
We now split the denominator and rewrite it as the product of 2 terms:
 $\Rightarrow \cos x \times \dfrac{{2\sin x}}{{\cos x \times \cos x}}$
Thus, we get:
$\Rightarrow \dfrac{{2\sin x}}{{\cos x}}$ …………………..(9)
Also, from trigonometric identities, we know that:
$\Rightarrow \dfrac{{\sin x}}{{\cos x}} = \tan x$ ……………………(10)
Thus, substituting the respective value from equation (10) in the expression (9), we finally have our expression as:
$ = 2\tan x$, which is nothing but the expression on the other side of the expression, that is, on the right side of the expression.
Hence, proved.

Note: The alternative way of solving the same problem would be to take a common denominator with $\tan x$ and $\sec x$ terms, without converting them into trigonometric ratios containing the $\sin x$ and $\cos x$ terms. Then, we will make use of the trigonometric identity $1 + {\tan ^2}x = {\sec ^2}x$. Further simplification is as usual.

To prove a trigonometric equation, we should tend to start with the more complicated side of the equation, and keep simplifying it until it is transformed into the same expression as on the other side of the given equation. In some cases, we can also try to simplify both sides of the equation and arrive at a common expression to prove their equality. The various procedures of solving a trigonometric equation are: expanding the expressions, making use of the identities, factoring the expressions or simply using basic algebraic strategies to obtain the desired results.