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  4. If x cos θ+y sin θ=p is the normal form and y=m x+c is the slope - intercept form of the line x+2 y+1=0, then textTan-1( tan θ+m+c)=

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

TS EAMCET 2020

Solution:

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