Question

How do you prove $2\tan \left( x \right)\sec \left( x \right) = \dfrac{1}{{1 - \sin \left( x \right)}} - \dfrac{1}{{1 + \sin \left( x \right)}}$?

Answer 1

Hint: To verify the given equation, we will at first, take the complicated side of the question, simplify it using trigonometric identities until it gets transformed into the same expression as on the other side of the equation. Moreover, some of the algebraic identities can also be used to simplify expressions.

Complete step-by-step solution:
We need to prove: $2\tan \left( x \right)\sec \left( x \right) = \dfrac{1}{{1 - \sin \left( x \right)}} - \dfrac{1}{{1 + \sin \left( x \right)}}$
Proof:
First, we consider the complex side, i.e., the RHS:
$\dfrac{1}{{1 - \sin \left( x \right)}} - \dfrac{1}{{1 + \sin \left( x \right)}}$
Solving further, we get:
$ = \dfrac{{1 + \sin \left( x \right) - \left[ {1 - \sin \left( x \right)} \right]}}{{\left[ {1 - \sin \left( x \right)} \right]\left[ {1 + \sin \left( x \right)} \right]}}$ ………………………(1)
We know that, $\left( {a - b} \right)\left( {a + b} \right) = {a^2} - {b^2}$
Thus, the denominator of the expression (1) can be rewritten as:
$\left[ {1 - \sin \left( x \right)} \right]\left[ {1 + \sin \left( x \right)} \right] = 1 - {\sin ^2}\left( x \right)$ ……………………….(2)
Therefore, after substituting the value of $\left[ {1 - \sin \left( x \right)} \right]\left[ {1 + \sin \left( x \right)} \right]$ from equation (2) in expression (1), we get:
$ = \dfrac{{1 + \sin \left( x \right) - \left[ {1 - \sin \left( x \right)} \right]}}{{1 - {{\sin }^2}\left( x \right)}}$
On simplifying further, we get:
$ = \dfrac{{1 + \sin \left( x \right) - 1 + \sin \left( x \right)}}{{1 - {{\sin }^2}\left( 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:
$ = \dfrac{{2\sin \left( x \right)}}{{1 - {{\sin }^2}\left( x \right)}}$ ………………………(3)
Now, we have a trigonometric identity as: ${\sin ^2}\left( x \right) + {\cos ^2}\left( x \right) = 1$ ……………………..(4)
On rearranging equation (4), we get:
$1 - {\sin ^2}\left( x \right) = {\cos ^2}\left( x \right)$ ……………………..(5)
Now, substituting the value of $\left[ {1 - {{\sin }^2}\left( x \right)} \right]$ from equation (5) in the expression (3), we get:
$ = \dfrac{{2\sin \left( x \right)}}{{{{\cos }^2}\left( x \right)}}$ ……………………(6)
We now split the denominator and rewrite it as the product of 2 terms:
 $ = \dfrac{{2\sin \left( x \right)}}{{\cos \left( x \right) \times \cos \left( x \right)}}$
On further splitting the above expression into 3 terms, we get:
$ = 2 \times \dfrac{{\sin \left( x \right)}}{{\cos \left( x \right)}} \times \dfrac{1}{{\cos \left( x \right)}}$ ………………….(7)
Also, from trigonometric identities, we know that:
$\dfrac{{\sin x}}{{\cos x}} = \tan x$ ……………………(8)
and $\dfrac{1}{{\cos x}} = \sec x$ …………………….(9)
Thus, substituting the respective values from equations (8) and (9) in the expression (7), we finally have our expression as:
=$2\tan \left( x \right)\sec \left( x \right)$, which is nothing but the expression on the other side of the expression, that is, on the left side of the expression.
Hence, proved.

Note: Here we have 2 basic identities to make the calculation easier. Remembering these identities helps us to solve the problem ${\sin ^2}\left( x \right) + {\cos ^2}\left( x \right) = 1$ and $\left( {a - b} \right)\left( {a + b} \right) = {a^2} - {b^2}$ .One of the most common practices of solving trigonometric equations involve using the algebraic identities. In fact, solving a trigonometric expression constantly involves implications of algebraic techniques. The approach of solving any trigonometric expression depends on the nature of the expression. However, it is always helpful to start from the complicated side of the equation.