Question

The most general value of $\theta $ satisfying the equations $\tan \theta =-1$ and $\cos \theta =\dfrac{1}{\sqrt{2}}$ is
A.$n\pi +\dfrac{7\pi }{4}$
B. $n\pi +{{(-1)}^{n}}\dfrac{7\pi }{4}$
C. $2n\pi +\dfrac{7\pi }{4}$
D. None of these.

Answer 1

Hint: To find the most general value of $\theta $ which will satisfy both the given equations we will first determine the quadrant in which $\theta $ belongs and then find its value accordingly with the help of trigonometric table of values. Using trigonometric table of values at specific angles for both equations we will form new equations and then apply the theorem of general solution of cos and tan.
For all the real values of $x$ and $y$, $\cos x=\cos y$ implies that $x=2n\pi \pm y$and for all the even multiples of $\dfrac{\pi }{2}$, $\tan x=\tan y$ implies that $x=2n\pi +y$ where $n\in Z$.

Complete step by step solution: We are given trigonometric equations $\tan \theta =-1$ and $\cos \theta =\dfrac{1}{\sqrt{2}}$ and we have to determine the most general value of $\theta $ which will satisfy both the equations.
From the given equations we can see that $\theta $ is negative for tan and positive for cos. It means that $\theta $ lies in the fourth quadrant where angle is represented as $2\pi -\theta $.
Now we know that $\tan \dfrac{\pi }{4}=1$ so,
$\begin{align}
  & \tan \theta =-\tan \dfrac{\pi }{4} \\
 & \tan \theta =\tan \left( 2\pi -\dfrac{\pi }{4} \right) \\
 & \tan \theta =\tan \left( \dfrac{7\pi }{4} \right)
\end{align}$
We will now apply the theorem of general equation of tan,
$\theta =2n\pi +\dfrac{7\pi }{4}$
We will now take $\cos \theta =\dfrac{1}{\sqrt{2}}$. As we know that $\cos \dfrac{\pi }{4}=\dfrac{1}{\sqrt{2}}$so,
$\begin{align}
  & \cos \theta =\cos \dfrac{\pi }{4} \\
 & \cos \theta =\cos \left( 2\pi -\dfrac{\pi }{4} \right) \\
 & \cos \theta =\cos \dfrac{7\pi }{4} \\
\end{align}$
Applying the theorem of general solution of cos, we will get,
$\theta =2n\pi \pm \dfrac{7\pi }{4}$ which means that there will be two general equation $\theta =2n\pi +\dfrac{7\pi }{4}$ and $\theta =2n\pi -\dfrac{7\pi }{4}$.
Now there are three values of $\theta $ satisfying each of the equations that is $\theta =2n\pi +\dfrac{7\pi }{4}$ , $\theta =2n\pi +\dfrac{7\pi }{4}$ and $\theta =2n\pi -\dfrac{7\pi }{4}$ but we have to find the most general equation so we will select $\theta =2n\pi +\dfrac{7\pi }{4}$ because it is common in both the general solutions of equation.
The most general value of $\theta $ satisfying the equations $\tan \theta =-1$ and $\cos \theta =\dfrac{1}{\sqrt{2}}$ is $\theta =2n\pi +\dfrac{7\pi }{4}$

Option ‘A’ is correct

Note: There are four quadrants in which trigonometric function is either positive or negative. In first quadrant all the functions are positive, in second quadrant only sin and cosec is positive, in third quadrant only tan and cot is positive, and in fourth quadrant only cos and sec is positive.