Question

Select the general solution(s) of the trigonometric equation ${\tan }^2\theta -\tan \theta =0.$

• A. $\theta =(2n+1)\frac{\pi }{2}\forall n\in Z$
• B. $\theta =n\pi +\frac{\pi }{4}\forall n\in Z$
• C. $\theta =n\pi +\frac{3\pi }{4}\forall n\in Z$
• D. $\theta =n\pi \forall n\in Z$

Answer 1

Answer: (B), (D)

$\theta =n\pi \forall n\in Z$

Given: ${\tan }^2\theta -\tan \theta =0$
$\Rightarrow \tan \theta (\tan \theta -1)=0$
$\Rightarrow \tan \theta =0\cdots \left(i\right)\text{or}\tan \theta =1\cdots \left(ii\right)$
Taking $\left(i\right),$ we get:
$\tan \theta =0\Rightarrow \theta =n\pi \cdots \left(A\right)$
Also, taking $\left(ii\right),$ we get:
$\tan \theta =1\Rightarrow \tan \theta =\tan \frac{\pi }{4}=\tan \frac{5\pi }{4}\Rightarrow \theta =n\pi +\frac{\pi }{4}\forall n\in Z\cdots \left(B\right)\&\theta =m\pi +\frac{5\pi }{4}\forall m\in Z\cdots \left(C\right)$