How do you solve: $\tan \theta + \sec \theta = 1$ ?

Hint: In this question we have to use the formula \[se{c^2}\theta - ta{n^2}\theta = 1\] to get to the final answer. In this formula we have to use a formula of algebra, that is ${a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right)$ . Then finally we will be left with two equations and two variables $\tan \theta \,\,and\,\,\sec \theta $ which we can easily find and then the value of $\theta $ .
Formula used: \[se{c^2}\theta - ta{n^2}\theta = 1\]

Complete step-by-step answer:
In the question, it is given that $\tan \theta + \sec \theta = 1 - - - - - \left( 1 \right)$.
Also, we know that \[se{c^2}\theta - ta{n^2}\theta = 1\].
Now using the formula ${a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right)$
$ \Rightarrow \left( {\tan \theta + \sec \theta } \right)\left( {\sec \theta - \tan \theta } \right) = 1$
On cross-multiplication, we get
$ \Rightarrow \sec \theta - \tan \theta = \dfrac{1}{{\tan \theta + \sec \theta }}$
Now using $\tan \theta + \sec \theta = 1$
$ \Rightarrow \sec \theta - \tan \theta = 1 - - - - - \left( 2 \right)$
Now, adding equation $\left( 1 \right)\,\,and\,\,\left( 2 \right)\,.$we get
$ \Rightarrow 2\sec \theta = 2$
On dividing both sides by $2$.
$ \Rightarrow \sec \theta = 1$
Now, using the identity $\sec \theta = \dfrac{1}{{\cos \theta }}$
$ \Rightarrow \dfrac{1}{{\cos \theta }} = 1$
Now doing the cross-multiplication,
$ \Rightarrow \cos \theta = 1$
We know that $\cos \theta $ takes the value $1$ infinite times as it is a periodic function with a period of $2\pi $ .
Therefore, $\theta = 2n\pi $ , where n is any integer.
So, the correct answer is “$\theta = 2n\pi $”.

Note: The Cos theta or cos θ is the ratio of the adjacent side to the hypotenuse, where θ is one of the acute angles. The cosine formula is as follows: $Cos \theta = \dfrac{Adjacent}{Hypotenuse}$. The general solution is given by $\theta = 2n\pi $