If $p =\frac{4\sin\theta \cos\theta}{\sin\theta +\cos\theta}$ Find the value of $\frac{p+2\sin\theta}{p-2\sin\theta}$ Problem : 

If $\displaystyle p =\frac{4\sin\theta\cos\theta}{\sin\theta +\cos\theta}$, find the value of $\displaystyle \frac{p+2\sin\theta}{p-2\sin\theta} + \frac{p+2\cos\theta}{p-2\cos\theta}$.

Please help how to proceed in such problem..Thanks..
 A: HINT:
So, we have $$\frac p{2\sin\theta}=\frac{2\cos\theta}{\sin\theta+\cos\theta}$$
Apply Componendo and dividendo to get $$\frac{p+2\sin\theta}{p-2\sin\theta}=\frac{2\cos\theta+\sin\theta+\cos\theta}{2\cos\theta-(\sin\theta+\cos\theta)}=\frac{\sin\theta+3\cos\theta}{\cos\theta-\sin\theta}$$
Similarly,  $$\frac p{2\cos\theta}=\frac{2\sin\theta}{\sin\theta+\cos\theta}\implies \frac{p+2\cos\theta}{p-2\cos\theta}=...=\frac{3\sin\theta+\cos\theta}{\sin\theta-\cos\theta}=-\frac{3\sin\theta+\cos\theta}{\cos\theta-\sin\theta}$$

Alternatively, $$\frac{p+2\sin\theta}{p-2\sin\theta}=\frac{\frac p{2\sin\theta}-1}{\frac p{2\sin\theta}+1}=\frac{\frac{2\cos\theta}{\sin\theta+\cos\theta}+1}{\frac{2\cos\theta}{\sin\theta+\cos\theta}-1}=\frac{2\cos\theta+(\sin\theta+\cos\theta)}{2\cos\theta-(\sin\theta+\cos\theta)}$$
$$\implies \frac{p+2\sin\theta}{p-2\sin\theta}=\frac{\sin\theta+3\cos\theta}{\cos\theta-\sin\theta}$$
Similarly for $$\frac{p+2\cos\theta}{p-2\cos\theta}$$
A: You can directly substitute the value P in the given term
We get:  
$\large \large \frac {3\sinθ  +  \cosθ}{\sinθ  -  \cosθ}    +         \frac {3\cosθ  +  \sinθ}{\cosθ -  \sinθ}$
which simplifies to $2$
A: Note that $$\frac{p}{2\cos\theta}=\frac{2\sin \theta}{\cos\theta+\sin\theta}$$ and 
$$\frac{p}{2\sin\theta}=\frac{2\cos \theta}{\cos\theta+\sin\theta}$$
Now apply the componendo-dividendo method to the first and second equalities and sum them up, you'll get the answer. It'll be $2$.
