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Question: Find the value \(\dfrac{{\sec \theta + \tan \theta - 1}}{{\tan \theta - \sec \theta + 1}}\) \( ...

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

Explanation

Solution

Hint: Here, we will use the trigonometric identity sec⁡2θ−tan⁡2θ=1{\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, sec⁡θ+tan⁡θ−1tan⁡θ−sec⁡θ+1\dfrac{{\sec \theta + \tan \theta - 1}}{{\tan \theta - \sec \theta + 1}}
AS, we know, the trigonometric identity sec⁡2θ−tan⁡2θ=1{\sec ^2}\theta - {\tan ^2}\theta = 1. So let us substitute the value of 1 in the numerator as sec⁡2θ−tan⁡2θ{\sec ^2}\theta - {\tan ^2}\theta , we get
⇒sec⁡θ+tan⁡θ−(sec⁡2θ−tan⁡2θ)tan⁡θ−sec⁡θ+1\Rightarrow \dfrac{{\sec \theta + \tan \theta - ({{\sec }^2}\theta - {{\tan }^2}\theta )}}{{\tan \theta - \sec \theta + 1}}
And also we know that sec⁡2θ−tan⁡2θ=(sec⁡θ+tan⁡θ)(sec⁡θ−tan⁡θ){\sec ^2}\theta - {\tan ^2}\theta = (\sec \theta + \tan \theta )(\sec \theta - \tan \theta ). Substituting it in the above formula, we get
⇒(sec⁡θ+tan⁡θ)−((sec⁡θ+tan⁡θ)(sec⁡θ−tan⁡θ))tan⁡θ−sec⁡θ+1\Rightarrow \dfrac{{(\sec \theta + \tan \theta ) - ((\sec \theta + \tan \theta )(\sec \theta - \tan \theta ))}}{{\tan \theta - \sec \theta + 1}}
Let us take common (sec⁡θ+tan⁡θ)(\sec \theta + \tan \theta )term in the numerator, we get
⇒(sec⁡θ+tan⁡θ)(1−(sec⁡θ−tan⁡θ))tan⁡θ−sec⁡θ+1 ⇒(sec⁡θ+tan⁡θ)(1−sec⁡θ+tan⁡θ)tan⁡θ−sec⁡θ+1  \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⁡θ+tan⁡θ)(1 - \sec \theta + \tan \theta )term gets cancel and we will be left with
⇒sec⁡θ+tan⁡θ\Rightarrow \sec \theta + \tan \theta
Hence, sec⁡θ+tan⁡θ−1tan⁡θ−sec⁡θ+1=sec⁡θ+tan⁡θ\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.