# Trigonometric equation 2

Solve the following equation. $$\left(\sqrt{\frac{1-\sin x}{1+\sin x}}-\sqrt{\frac{1+\sin x}{1-\sin x}}\right)\left(\sqrt{\frac{1-\cos x}{1+\cos x}}-\sqrt{\frac{1+\cos x}{1-\cos x}}\right)=-4.$$

I tried amplifying the first factor with $$\sqrt{\frac{1-\sin x}{1+\sin x}}+\sqrt{\frac{1+\sin x}{1-\sin x}}$$ and the second one by $$\sqrt{\frac{1-\cos x}{1+\cos x}}+\sqrt{\frac{1+\cos x}{1-\cos x}}$$this made some simplifications but still not enough.

• Hint : Rationalize each of the fraction might simplify.. – Ash Pd Apr 13 '17 at 18:40

The first factor is \begin{eqnarray} \sqrt{\dfrac{1-\sin x}{1+\sin x}}-\sqrt{\dfrac{1+\sin x}{1-\sin x}}&=&\dfrac{(\sqrt{1-\sin x})^2-(\sqrt{1+\sin x})^2}{\sqrt{(1+\sin x)(1-\sin x)}}\\ &=&\dfrac{(1-\sin x)-(1+\sin x)}{\sqrt{1-\sin^2x}}\\ &=&\dfrac{-2\sin x}{\sqrt{\cos^2x}}\\ &=&\dfrac{-2\sin x}{|\cos x|} \end{eqnarray} while the second factor is \begin{eqnarray} \sqrt{\dfrac{1-\cos x}{1+\cos x}}-\sqrt{\dfrac{1+\cos x}{1-\cos x}}&=&\dfrac{(\sqrt{1-\cos x})^2-(\sqrt{1+\cos x})^2}{\sqrt{(1+\cos x)(1-\cos x)}}\\ &=&\dfrac{(1-\cos x)-(1+\cos x)}{\sqrt{1-\cos^2x}}\\ &=&\dfrac{-2\cos x}{\sqrt{\sin^2x}}\\ &=&\dfrac{-2\cos x}{|\sin x|} \end{eqnarray} The original equation becomes then \begin{eqnarray} \dfrac{-2\sin x}{|\cos x|}\cdot\dfrac{-2\cos x}{|\sin x|}&=&-4\\ \dfrac{4\sin x\cos x}{|\sin x\cos x|}&=&-4\\ |\sin x\cos x|&=&-\sin x\cos x\\ |\sin2x|&=&-\sin 2x \end{eqnarray} It follows that $$\pi+2k\pi \le 2x\le 2\pi+2k\pi, \quad k\in \mathbb{Z},$$ i.e. $$\dfrac{\pi}{2}+k\pi \le x\le \pi+k\pi, \quad k \in \mathbb{Z}.$$

We need to solve $$\frac{(1-\sin{x}-1-\sin{x})(1-\cos{x}-1-\cos{x})}{|\sin{x}\cos{x}|}=-4$$ or $$|\sin{x}\cos{x}|=-\sin{x}\cos{x}$$ or $$\sin2x<0$$ or $\frac{\pi}{2}+\pi k<x<\pi+\pi k$, where $k\in\mathbb Z$.

Hint:

Observe that $$\sqrt{\frac{1-\sin x}{1+\sin x}}-\sqrt{\frac{1+\sin x}{1-\sin x}}=\frac{1-\sin x-(1+\sin x)}{\sqrt{1-\sin^2 x}}=\frac{-2\sin x}{|\cos x|}$$ and

$$\sqrt{\frac{1-\cos x}{1+\cos x}}-\sqrt{\frac{1+\cos x}{1-\cos x}}=\frac{1-\cos x-(1+\cos x)}{\sqrt{1-\cos^2 x}}=\frac{-2\cos x}{|\sin x|}$$ Then, the given equation is equivalent to $$\frac{4\sin x\cos x}{|\cos x \sin x|}=-4$$

• Provided that $\sin{x}\cos{x}<0$. – Chappers Apr 13 '17 at 18:46

i will use change of variable(using the unit circle $x = \cos t, \ y = \sin t, \ x^2 + y^2 = 1$), you want $$\left(\sqrt{\frac{1-y}{1+y}}-\sqrt{\frac{1+y}{1-y}}\right)\left(\sqrt{\frac{1-x}{1+ x}}-\sqrt{\frac{1+x}{1-x}}\right)= \left(\frac{1-y-(1+y)}{\sqrt{1-y^2}}\right) \left(\frac{1-x-(1+x)}{\sqrt{1-x^2}}\right)\\ =\frac{(-2y)}{|x|}\frac{(-2x)}{|y|}=\pm4.$$