Question

Match the options in List 1 with options in List 2

Answer 1

A) If $\cos B-\sin A>0$ then $\sin B-\cos A<0$
And if $\cos B-\sin A<0$ then $\sin B-\cos A>0$
Hence the point is in $2nd$ or $4th$ quadrant

B) As $2^{\sin \theta }>1\Rightarrow \sin \theta >{\log }_{2}1\Rightarrow \sin \theta >0$
and $3^{\cos \theta }<1\Rightarrow \cos \theta <{\log }_{3}1\Rightarrow \cos \theta <0$
Therefore, $\theta$ lies in $2nd$quadrant

C) From triangle inequality $|\cos x+\sin x|\ge \left|\sin x\right|+\left|\cos x\right|$
and equality holds when $\sin x.\cos x>0$
Therefore x lies in $1st$ or $3rd$ quadrant

D) $\sqrt{\frac{1-\sin A}{1+\sin A}}+\frac{\sin A}{\cos A}=\sqrt{\frac{1-\sin A}{1+\sin A}\times \frac{1-\sin A}{1-\sin A}}+\frac{\sin A}{\cos A}$
$=\frac{\sqrt{{(1-\sin A)}^2}}{\sqrt{1-{\sin }^{2}A}}+\frac{\sin A}{\cos A}$ ...(1)
If A is in $1st$ quadrant then (1) gives
$\frac{1-\sin A}{\cos A}+\frac{\sin A}{\cos A}=\frac{1}{\cos A}$
And if A is in $4th$ quadrant then (1) gives
$\frac{(1-\sin A)}{\cos A}+\frac{\sin A}{\cos A}=\frac{1}{\cos A}$

In other cases the equation will not hold.