Question

Evaluate $\dfrac{{1 + \sin x}}{{1 - \sin x}}$

Answer 1

The given function is trigonometric with respect to sine function, in which to evaluate $\dfrac{{1 + \sin x}}{{1 - \sin x}}$, such that we must know all the basic trigonometric relations i.e., formulas related with sin functions and apply the formula$\left( {a - b} \right)\left( {a + b} \right) = \left( {{a^2} - {b^2}} \right)$ and other functions.
Formula used:
$\left( {a - b} \right)\left( {a + b} \right) = \left( {{a^2} - {b^2}} \right)$

Complete step by step answer:
Given,
$\dfrac{{1 + \sin x}}{{1 - \sin x}}$
Now, multiply and divide the given expression by $1 + \sin x$, as:
$\Rightarrow \dfrac{{\left( {1 + \sin x} \right)\left( {1 + \sin x} \right)}}{{\left( {1 - \sin x} \right)\left( {1 + \sin x} \right)}}$
Here, $\left( {1 - \sin x} \right)\left( {1 + \sin x} \right)$ can be expressed as $\left( {1^2 - \sin^2 x} \right)$, using the identity $\left( {a - b} \right)\left( {a + b} \right) = \left( {{a^2} - {b^2}} \right)$, hence we get:
$= \dfrac{{{{\left( {1 + \sin x} \right)}^2}}}{{\left( {1^2 - \sin^2 x} \right)}}$
We know that, ${\sin ^2}x + {\cos ^2}x = 1$ and $1 - {\cos ^2}x = \sin^2x$, hence we get:
$= \dfrac{{{{\left( {1 + \sin x} \right)}^2}}}{{{{\cos }^2}x}}$
= {\left\\{ {\dfrac{{\left( {1 + \sin x} \right)}}{{\cos x}}} \right\\}^2}
= {\left\\{ {\sec x + \tan x} \right\\}^2}
Therefore,
\dfrac{{1 + \sin x}}{{1 - \sin x}} = {\left\\{ {\sec x + \tan x} \right\\}^2}

Note: The key point to solve any trigonometric function is that we must know all the formulas with respect to the related questions asked as it seems easy to solve the question, we must note the chart of all related functions with respect to the equation, and here are some of the formulas to be noted while solving:
${\sin ^2}\theta + {\cos ^2}\theta = 1$ , ${\tan ^2}\theta + 1 = {\sec ^2}\theta$