 QUESTION

# Solve the equation given below to find the value of $\theta$.$\sqrt{2}\sec \theta +\tan \theta =1$

Hint: This question can be solved by applying trigonometric formulas that we have studied in the chapter trigonometry. Convert all the trigonometric functions that are given in the question into $\sin \theta$ and $\cos \theta$ and then, use the formula $\cos x\cos y-\sin x\sin y=\cos \left( x+y \right)$. Then, find the general solution of the obtained equation. Using this, we can solve this question.

Before proceeding with the question, we must know all the formulas that will be required to solve this question.
In trigonometry, we can convert the tan and the sec functions to sin and cos functions by using the formulas,
$\tan \theta =\dfrac{\sin \theta }{\cos \theta }$ . . . . . . . . . . . . (1)
$\sec \theta =\dfrac{1}{\cos \theta }$ . . . . . . . . . . . . (2)
Also, from trigonometry, the general solution of the equation $\cos \theta =m$ is,
$\theta =2n\pi \pm \left( Principal\text{ solution }of~{{\cos }^{-1}}m \right)$ . . . . . . . . . . . . (3)
Also, in trigonometry, we have a formula $\cos x\cos y-\sin x\sin y=\cos \left( x+y \right)$ . . . . . . . . . . (4)
In the question, we are given an equation $\sqrt{2}\sec \theta +\tan \theta =1$ and we have to solve this equation to find the value of $\theta$.
Using formula (1) and (2), the equation given in the question can be also written as,
\begin{align} & \dfrac{\sqrt{2}}{\cos \theta }+\dfrac{\sin \theta }{\cos \theta }=1 \\ & \Rightarrow \dfrac{\sqrt{2}+\sin \theta }{\cos \theta }=1 \\ & \Rightarrow \sqrt{2}+\sin \theta =\cos \theta \\ & \Rightarrow \cos \theta -\sin \theta =\sqrt{2} \\ & \Rightarrow \dfrac{1}{\sqrt{2}}\cos \theta -\dfrac{1}{\sqrt{2}}\sin \theta =1 \\ \end{align}
In trigonometry, we have $\sin \dfrac{\pi }{4}=\cos \dfrac{\pi }{4}=\dfrac{1}{\sqrt{2}}$. Using this in the above equation, we get,
$\cos \dfrac{\pi }{4}\cos \theta -\sin \dfrac{\pi }{4}\sin \theta =1$
Using formula (4) in the above equation, we can write the above equation as,
$\cos \left( \theta +\dfrac{\pi }{4} \right)=1$
From formula (3), by substituting m = -1, the general solution of the above equation is given by,
\begin{align} & \theta +\dfrac{\pi }{4}=2n\pi \pm \left( Principal\text{ solution }of~{{\cos }^{-1}}1 \right) \\ & \Rightarrow \theta +\dfrac{\pi }{4}=2n\pi \pm \left( 0 \right) \\ & \Rightarrow \theta =2n\pi -\dfrac{\pi }{4} \\ \end{align}
Where n is an integer.
Hence, the answer is $\theta =2n\pi -\dfrac{\pi }{4}$ where, n is an integer.

Note: There is a possibility that one may commit a mistake while using formula (4). It is a common mistake of writing this formula as $\cos x\cos y-\sin x\sin y=\cos \left( x-y \right)$ instead of the correct formula $\cos x\cos y-\sin x\sin y=\cos \left( x+y \right)$ which will lead us to an incorrect answer.A solution of a trigonometric equation is the value of the unknown angle that satisfies the equation. Example: Solving an equation means to find the set of all values of the unknown value which satisfy the given equation. The solutions lying between 0 to 2π or between 0° to 360° are called principal solutions.Students should know the general and principal solutions of trigonometric functions for solving these types of problems.