Question

Verify the following equation
\[4\tan x\sec x = \dfrac{{\left( {1 + \sin x} \right)}}{{\left( {1 - \sin x} \right)}} - \dfrac{{\left( {1 - \sin x} \right)}}{{\left( {1 + \sin x} \right)}}\]

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 have,
\[4\tan x\sec x = \dfrac{{\left( {1 + \sin x} \right)}}{{\left( {1 - \sin x} \right)}} - \dfrac{{\left( {1 - \sin x} \right)}}{{\left( {1 + \sin x} \right)}}\]
Taking the complicate part to solve, that is, RHS
 \[ \Rightarrow \dfrac{{\left( {1 + \sin x} \right)}}{{\left( {1 - \sin x} \right)}} - \dfrac{{\left( {1 - \sin x} \right)}}{{\left( {1 + \sin x} \right)}} - - - - - \left( 1 \right)\]
Taking L.C.M. of the denominator,
We get
\[ \Rightarrow \dfrac{{\left( {1 + \sin x} \right)\left( {1 + \sin x} \right) - \left( {1 - \sin x} \right)\left( {1 - \sin x} \right)}}{{\left( {1 - \sin x} \right)\left( {1 + \sin x} \right)}}\]
Further simplifying,
\[ \Rightarrow \dfrac{{{{\left( {1 + \sin x} \right)}^2} - {{\left( {1 - \sin x} \right)}^2}}}{{\left( {1 - \sin x} \right)\left( {1 + \sin x} \right)}} - - - - - \left( 2 \right)\]
For solving the brackets in numerator of the (2) equation, we will use the most commonly used algebraic identities. They are as follows:
\[{\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2}\] And
\[{\left( {a - b} \right)^2} = {a^2} - 2ab + {b^2}\]
Comparing and putting the values from the identities into the equation (2)
We get,
\[ \Rightarrow \dfrac{{\left( {1 + 2\sin x + {{\sin }^2}x} \right) - \left( {1 - 2\sin x + {{\sin }^2}x} \right)}}{{\left( {1 - \sin x} \right)\left( {1 + \sin x} \right)}}\]
For simplifying the denominator we will use the algebraic identity of difference of squares, which is
\[\left( {{a^2} - {b^2}} \right) = (a - b)\left( {a + b} \right)\]
Therefore, after comparing and putting the vale in the equation, we get the whole equation as,
\[ \Rightarrow \dfrac{{\left( {1 + 2\sin x + {{\sin }^2}x} \right) - \left( {1 - 2\sin x + {{\sin }^2}x} \right)}}{{\left( {1 - {{\sin }^2}x} \right)}}\]
Therefore,
\[ \Rightarrow \dfrac{{\left( {1 + 1} \right)\left( {2\sin x} \right)}}{{\left( {1 - {{\sin }^2}x} \right)}} - - - - - \left( 3 \right)\]
Now, according to trigonometric identities, we know that
\[ \Rightarrow {\sin ^2}x + {\cos ^2}x = 1\]
\[ \Rightarrow 1 - si{n^2}x = {\cos ^2}x - - - - - \left( a \right)\]
Thus, after substituting the value of (a) in the denominator of (3), we get,
\[ \Rightarrow \dfrac{{4\sin x}}{{{{\cos }^2}x}}\]
Which can be rewritten as,
\[ \Rightarrow 4 \times \dfrac{{\sin x}}{{\cos x}} \times \dfrac{1}{{\cos x}} - - - - - \left( 4 \right)\]
From the quotient and reciprocal trigonometric identities, we know that,
\[\dfrac{{\sin x}}{{\cos x}} = \tan x\] And \[\dfrac{1}{{\cos x}} = \sec x\]
Thus, after substituting these values in equation (4) we get,
\[ \Rightarrow 4\tan x\sec x = LHS\]
Hence, verified.

Note: 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.