Question

If $\sin \left(\theta \right)+\cos \left(\theta \right)=0$ and $0<\theta <\pi$, then $\theta$ is equal to

• A. $0$
• B. $\frac{\pi }{4}$
• C. $\frac{\pi }{2}$
• D. $\frac{3\pi }{4}$

Answer 1

The correct option is D

$\frac{3\pi }{4}$

Solve the given function for $\theta$:

From the given equation, we can conclude that

$\sin \left(\theta \right)=-\cos \left(\theta \right)$

Divide both the sides by $\cos \left(\theta \right)$,

$\begin{array}{rcl}\therefore \frac{\sin \left(\theta \right)}{\cos \left(\theta \right)}& =& -\frac{\cos \left(\theta \right)}{\cos \left(\theta \right)}\\ & \Rightarrow & \tan \left(\theta \right)=-1\end{array}$

With the given condition $0<\theta <\pi$, we know that $\tan$ is negative in the second quadrant.

Then, the only viable angle is $\frac{3\pi }{4}$.

Hence, the correct option is (D).