COLUMN-I		COLUMN-II
a)	Suppose ABC is a triangle with three acute  angles A,B, and C. The point whose coordinates  are ($\cos B-\sin A,\sin B-CosA$) can be in the	p)	$1^{st}$ quadrant
b)	If $2^{\sin \theta }>1and{3}^{\cos \theta }<1,then\theta \in$	q)	2nd quadrant
c)	For $|\cos x+\sin x|=\left|\sin x\right|+\left|\cos x\right|,$  X belongs to	r)	$3^{rd}$ quadrant
d)	If $\sqrt{\frac{1-\sin A}{1+\sin A}}+\frac{\sin A}{\cos A}=\frac{1}{\cos A},$  for all permissible values of A,  then A can belong to	s)	$4^{th}$ quadrant

a. Since angles, A, B and C are acute angles, we have 
$A+B>\pi /2$ 
$A>\frac{\pi }{2}-B$ 
$SinA-\cos B>0$ 
$\Rightarrow \cos B-\sin A<0$ -------(1)
Again, $B>\frac{\pi }{2}-A$ 
$\sin B>CosA$ 
$\sin B-CosA>0$ -----------(2) 
From 1 and 2  point is in second quadrant 
b. $2^{\sin \theta }>1\Rightarrow \sin \theta >0\Rightarrow \theta \in 1^{st}or{2}^{nd}$  quadrant 
$3^{\cos \theta }<1\Rightarrow \cos \theta <0\Rightarrow \theta \in 2^{nd}or{3}^{rd}$  quadrant
Hence, $\theta \in 2^{nd}$ quadrant 
c. $|\cos x+\sin x|=\left|\sin x\right|+\left|\cos x\right|$ Thus, $\cos x$ and $\sin x$ must have same sing or at least one is zero 
So, $x\in 2^{nd}or{4}^{th}$ quadrant 
d. $L.H.S=\frac{1-\sin A}{\left|\cos A\right|}+\frac{\sin A}{\cos A}=\frac{1}{\cos A}$ 
Which is true only if $\left|CosA\right|=\cos A$