Question

Find the value $\dfrac{{\sec \theta + \tan \theta - 1}}{{\tan \theta - \sec \theta + 1}}$
$
  A)\sec \theta - \tan \theta \\
  B)\tan \theta - \sec \theta \\
  C)\sec \theta + \tan \theta \\
  D)1 \\
 $

Answer 1

Hint: Here, we will use the trigonometric identity ${\sec ^2}\theta - {\tan ^2}\theta = 1 $to simplify the given equation and then by applying the simple formulae and operations the value is calculated.

Complete step-by-step answer:
Given, $\dfrac{{\sec \theta + \tan \theta - 1}}{{\tan \theta - \sec \theta + 1}}$
AS, we know, the trigonometric identity ${\sec ^2}\theta - {\tan ^2}\theta = 1$. So let us substitute the value of 1 in the numerator as ${\sec ^2}\theta - {\tan ^2}\theta $, we get
$ \Rightarrow \dfrac{{\sec \theta + \tan \theta - ({{\sec }^2}\theta - {{\tan }^2}\theta )}}{{\tan \theta - \sec \theta + 1}}$
And also we know that ${\sec ^2}\theta - {\tan ^2}\theta = (\sec \theta + \tan \theta )(\sec \theta - \tan \theta )$. Substituting it in the above formula, we get
$ \Rightarrow \dfrac{{(\sec \theta + \tan \theta ) - ((\sec \theta + \tan \theta )(\sec \theta - \tan \theta ))}}{{\tan \theta - \sec \theta + 1}}$
Let us take common $(\sec \theta + \tan \theta )$term in the numerator, we get
$
   \Rightarrow \dfrac{{(\sec \theta + \tan \theta )(1 - (\sec \theta - \tan \theta ))}}{{\tan \theta - \sec \theta + 1}} \\
   \Rightarrow \dfrac{{(\sec \theta + \tan \theta )(1 - \sec \theta + \tan \theta )}}{{\tan \theta - \sec \theta + 1}} \\
 $
Now, from the above equation, $(1 - \sec \theta + \tan \theta )$term gets cancel and we will be left with
$ \Rightarrow \sec \theta + \tan \theta $
Hence, $\dfrac{{\sec \theta + \tan \theta - 1}}{{\tan \theta - \sec \theta + 1}} = \sec \theta + \tan \theta $.
So, option C is the correct option.

Note: To solve the given problem we need to have basic knowledge about trigonometry chapter. To solve the problem we need to think about converting values and trigonometry formulas which will be helpful for us.