Question

If the slope of the line passing through the points $(2,\sin \theta )$ and $(1,\cos \theta )$ is $0,$ then the general solution of $\theta$, is

• A. $n\pi +\frac{\pi }{4},\forall n\in Z$
• B. $n\pi -\frac{\pi }{4},\forall n\in Z$
• C. $n\pi \pm \frac{\pi }{4},\forall n\in Z$
• D. $n\pi ,\forall n\in Z$

Answer 1

Answer: (A)

$n\pi +\frac{\pi }{4},\forall n\in Z$

Line passing through points $(2,\sin \theta )$ and $(1,\cos \theta )$ is
$y-\cos \theta =(\sin \theta -\cos \theta )(x-1)$
$y=(\sin \theta -\cos \theta )x+2\cos \theta -sin\theta$
Slope$=\sin \theta -\cos \theta =0$ (given)
$\sin \theta =\cos \theta$
$\tan \theta =1$
$\theta =n\pi +\frac{\pi }{4},\forall n\in Z$

Hence, option A is the correct answer.