Question

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

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

Answer 1

The correct option is C $\frac{79}{18}$

$\sin p+\cos p=\frac{1}{2}\Rightarrow {(\sin p+\cos p)}^2=\frac{1}{4}1+2\sin p.\cos p=\frac{1}{4}\Rightarrow \sin p\cos p=-\frac{3}{8}$

Therefore,

$x+y=\frac{{\sin }^{3}p}{{\cos }^{2}p}+\frac{{\cos }^{3}p}{{\sin }^{2}p}$

$=\frac{{\sin }^{5}p+{\cos }^{5}p}{{\sin }^{2}p{\cos }^{2}p}$

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

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

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

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

$=\frac{{\left(\frac{1}{2}\right)}^3-3(-\frac{3}{8})\left(\frac{1}{2}\right)-\left(\frac{9}{64}\right)\left(\frac{1}{2}\right)}{\left(\frac{9}{64}\right)}$

$=\frac{\frac{1}{8}+\frac{9}{16}-\frac{9}{128}}{\frac{9}{64}}=\frac{79}{18}$