Question

If $x=\frac{{\sin }^{3}p}{{\cos }^{2}p},y=\frac{{\cos }^{3}p}{{\sin }^{2}p}$ and $\sin p+\cos p=\frac{1}{2}$, then $x+y$ is equal to

• A. $\frac{75}{18}$
• B. $\frac{44}{9}$
• C. $\frac{79}{18}$
• D. $\frac{48}{9}$

Answer 1

Answer: (C)

$\frac{79}{18}$

Given $x=\frac{{\sin }^{3}p}{{\cos }^{2}p},y=\frac{{\cos }^{3}p}{{\sin }^{2}p}$; and $\sin p+\cos p=\frac{1}{2}$

Consider, $x+y=\frac{{\sin }^{3}p}{{\cos }^{2}p}+\frac{{\cos }^{3}p}{{\sin }^{2}p}$
$\Rightarrow x+y=\frac{{\sin }^{5}p+{\cos }^{5}p}{{\sin }^{2}p{\cos }^{2}p}$

$=\frac{{\sin }^{3}p(1-{\cos }^{2}p)+{\cos }^{3}p(1-{\sin }^{2}p)}{{\sin }^{2}p{\cos }^{2}p}$

$=\frac{{\sin }^{3}p+{\cos }^{3}p-{\sin }^{2}p{\cos }^{2}p(\sin p+\cos p)}{{\sin }^{2}p{\cos }^{2}p}$
$=\frac{{(\sin p+\cos p)}^3-3\sin p\cos p(\sin p+\cos p)-{\sin }^{2}p{\cos }^{2}p(\sin p+\cos p)}{{\sin }^{2}p{\cos }^{2}p}$ .....(1)
Since $(\sin p+cosp{)}^2={\sin }^{2}p+{\cos }^{2}p+2\sin p\cos p$
$\Rightarrow \sin p\cos p=-\frac{3}{4}$
So, using in (1),
$x+y=\frac{\frac{1}{8}+\frac{9}{16}-\frac{9}{32}}{\frac{9}{16}}=\frac{79}{18}$