Question

If $x\cos \theta +y\sin \theta =p$ is the normal form and $y=mx+c$ is the slope - intercept form of the line $x+2y+1=0$, then ${\text{Tan}}^{-1}(\tan \theta +m+c)=$

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

Answer 1

Answer: (D)

We have, $x+2y+1=0$
$\Rightarrow x+2y=-1$
$\Rightarrow -x-2y=1$
$\Rightarrow \frac{-1}{\sqrt{5}}x-\frac{2}{\sqrt{5}}y=\frac{1}{\sqrt{5}}$
$\therefore P=\frac{1}{\sqrt{5}},\cos \theta =\frac{-1}{\sqrt{5}},\sin \theta =\frac{-2}{\sqrt{5}}$
$\therefore \tan \theta =\frac{\sin \theta }{\cos \theta }=2$
Now, again
$x+2y+1=0$
$\Rightarrow 2y=-x-1$
$\Rightarrow y=\frac{-1}{2}x-\frac{1}{2}$
$\therefore m=\frac{-1}{2},c=\frac{-1}{2}$
$\therefore {\tan }^{-1}(\tan \theta +m+c)$
$={\tan }^{-1}\left(2-\frac{1}{2}-\frac{1}{2}\right)$
$={\tan }^{-1}(1)=\frac{\pi }{4}$