If $\mathop {\lim }\limits_{x \to 0}\left(\frac{p \sin 2 x+1-\cos 2 x}{x+\tan x}\right)=1$ then the value of ' $p$ ' is
$2$
$-1$
$1$
$\frac{1}{2}$
Let the line $L_1$ be a line passing through the point $(\mathbf{0},-\mathbf{6})$ and making an angle of $\mathbf{1 5 0}^{\circ}$ with the positive $x$-axis. Then the equation of a line $L_2$ parallel to $L_1$ and crossing the $y$-axis 2 units below the origin is:
$$ x \sqrt{3}+y+6=0 $$
$$ x-\sqrt{3} y+6 \sqrt{3}=0 $$
$$ x-\sqrt{3} y-2 \sqrt{3}=0 $$
$$ x+\sqrt{3} y+2 \sqrt{3}=0 $$
$$ \mathop {\lim }\limits_{x \to {\pi \over 2}}\left(\frac{1-\sin x}{\cos x}\right) \text { is equal to } $$
$1$
$-1$
$\frac{1}{2}$
$0$
If $X=\tan ^{-1}\left[2 \cos \left(2 \sin ^{-1} \frac{1}{2}\right)\right]+\cos ^{-1}\left[\cos \left(\frac{7 \pi}{6}\right)\right]$ and $Y=\sin ^{-1}\left[\sin \left(\frac{11 \pi}{6}\right)\right]+\tan ^{-1}\left[\tan \left(\frac{4 \pi}{3}\right)\right]$ then the value of $\mathbf{2} \boldsymbol{X}-\boldsymbol{Y}$ is:
$\frac{3 \pi}{2}$
$2 \pi$
$ 0$
$\pi$
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