Question

The general value of $\theta$ is the equation $\cos \theta =\frac{1}{\sqrt{2}}$, $\tan \theta =-1$ is

• A. $2n\mathrm{\pi }+\frac{\mathrm{\pi }}{6},\mathrm{n}\in \mathrm{I}$
• B. $2n\mathrm{\pi }+\frac{7\mathrm{\pi }}{4},\mathrm{n}\in \mathrm{I}$
• C. $2n\mathrm{\pi }+{\left(-1\right)}^n\frac{\mathrm{\pi }}{3},\mathrm{n}\in \mathrm{I}$
• D. $2n\mathrm{\pi }+{\left(-1\right)}^n\frac{\mathrm{\pi }}{4},\mathrm{n}\in \mathrm{I}$

Answer 1

The correct option is B

$2n\mathrm{\pi }+\frac{7\mathrm{\pi }}{4},\mathrm{n}\in \mathrm{I}$

Explanation for the correct option:

Find the general value of $\theta$.

It is given that, $\cos \theta =\frac{1}{\sqrt{2}}$ and $\tan \theta =-1$.

Since, $\cos \theta$ is positive and $\tan \theta$ is negative. which implies that $\theta$ is in the fourth quadrant.

From $\cos \theta =\frac{1}{\sqrt{2}}$.

$$\begin{gathered} \cos \theta =\cos \left(\frac{\mathrm{\pi }}{4}\right) \\ \Rightarrow \cos \theta =\cos \left(2\mathrm{\pi }-\frac{\mathrm{\pi }}{4}\right) \\ \Rightarrow \cos \theta =\cos \left(\frac{7\mathrm{\pi }}{4}\right) \end{gathered}$$

Therefore, the value of $\theta$ in the fourth quadrant is $\frac{7\mathrm{\pi }}{4}$.

Similarly, $\tan \theta =-1$.

$$\begin{gathered} \tan \theta =-\tan \left(\frac{\mathrm{\pi }}{4}\right) \\ \Rightarrow \tan \theta =\tan \left(2\mathrm{\pi }-\frac{\mathrm{\pi }}{4}\right) \\ \Rightarrow \tan \theta =\tan \left(\frac{7\mathrm{\pi }}{4}\right) \end{gathered}$$

Therefore, the value of $\theta$ in the fourth quadrant is $\frac{7\mathrm{\pi }}{4}$.

So, the general value of $\theta$ can be given by $2n\mathrm{\pi }+\frac{7\mathrm{\pi }}{4},\mathrm{n}\in \mathrm{I}$.

Hence, Option $B$ is the correct answer.