If $\sin y+\cos y=\frac{1}{2}$ then find $\frac{\sin^3y}{\cos^2y}+\frac{\cos^3y}{\sin^2y}$ 
If $$\sin y+\cos y=\frac{1}{2} \tag{1}$$ Then find $$x=\frac{\sin^3y}{\cos^2y}+\frac{\cos^3y}{\sin^2y} \tag{2}$$

Given that $$\sin y=\frac{1}{2}-\cos y$$
Squaring both sides we get
$$8\cos^2y-4\cos y-3=0$$ 
Hence
$$\cos y=\frac{1-\sqrt{7}}{4}$$ and so $$\sin y=\frac{1+\sqrt{7}}{4}$$ 
substituting the above values in Eq $(2)$ and using the Binomial Theorem we get
$$x=\frac{\sin^5y+\cos^5y}{\sin^2y \cos^2y}$$ so
$$x=\frac{\dfrac{\left(1-\sqrt{7}\right)^5+\left(1+\sqrt{7}\right)^5}{4^5}}{\dfrac{36}{256}}$$ 
Now by the Binomial theorem $$\left(1-\sqrt{7}\right)^5+\left(1+\sqrt{7}\right)^5=2\left(1+10 \times 7+5 \times 49\right)=632$$
So $$x=\frac{\dfrac{632}{4^5}}{\dfrac{36}{256}}=\frac{79}{18}$$
I feel this is a very lengthy approach; can I get a better approach?
 A: 
Can i get any better approach

How about the following way?
Let $s=\sin y,c=\cos y$.
Squaring the both sides of $s+c=1/2$ gives
$$s^2+2sc+c^2=\frac 14\quad\Rightarrow\quad sc=-\frac 38.$$
Hence,
$$\begin{align}\frac{s^3}{c^2}+\frac{c^3}{s^2}&=\frac{s^5+c^5}{(sc)^2}\\\\&=\frac{(s^2+c^2)(s^3+c^3)-s^2c^2(s+c)}{(sc)^2}\\\\&=\frac{s^3+c^3-(sc)^2/2}{(sc)^2}\\\\&=\frac{(s+c)(s^2-sc+c^2)-(sc)^2/2}{(sc)^2}\\\\&=\frac{(1/2)(1-sc)-(sc)^2/2}{(sc)^2}\\\\&=\frac{79}{18}\end{align}$$
A: Write 
$$
x = s (s^2/c^2) + c (c^2/s^2) 
$$
then replace $c^2 = 1-s^2$ and vice versa to get
$$
x = s (1-c^2)/c^2 + c (1-s^2)/s^2 = (s/c^2) - s + (c/s^2) - c
$$
so that
$$
x + s + c = \frac{s}{c^2} + \frac{c}{s^2} = \frac{sc^2 + cs^2}{c^2s^2} = \frac{c + s}{cs}
$$ 
Since $c + s = 1/2$, this gives
$$
x + \frac{1}{2} = \frac{\frac{1}{2}}{cs}
$$
so 
$$
2x + 1 = \frac{1}{cs}
$$
Now 
$$
(c+s) = 1/2 \\
(c+s)^2 = 1/4 \\
c^2 + 2cs + s^2 = 1/4\\
2cs + 1 = 1/4\\
2cs=  -3/4 \\
cs = -3/8
$$
so the formula above becomes
$$
2x + 1 = \frac{1}{-3/8} = -\frac{8}{3}
$$
which you can solve. 
