# Problem 4. (3 points)

Find $\operatorname{tg} x + \operatorname{ctg} x$, given that $\sec x - \operatorname{cosec} x = 4 \sqrt{3}$.

Recall that $\sec x = 1 / \cos x$, and $\operatorname{cosec} x = 1 / \sin x$
Answer: -6 or 8.

Solution:

We will use the following formulas: $\operatorname{tg} x+\operatorname{ctg} x=\frac{1}{\sin x \cos x} ; \sec x-\operatorname{cosec} x=$ $\frac{\sin x-\cos x}{\sin x \cos x}$.

Introduce the notations: $t=\sin x \cos x$ and $A=\sec x-\operatorname{cosec} x$.

Then $A^{2}=\frac{\cos ^{2} x-2 \cos x \sin x+\sin ^{2} x}{t^{2}}=\frac{1-2 t}{t^{2}}$. Form the quadratic equation $1-2 t-t^{2} A^{2}=0$ and solve it for $t$:

$$
t=\frac{-1 \pm \sqrt{1+A^{2}}}{A^{2}}
$$

For $A=4 \sqrt{3}=\sqrt{48}$, we get $t=-\frac{8}{48}=-\frac{1}{6}$ and $t=\frac{6}{48}=\frac{1}{8}$. In the answer, we get the reciprocals of $t$.

It remains to be convinced that for each of the obtained values, there exists an $x$ for which this value is achieved for the given $A$.

Indeed, $t=\frac{1}{2} \sin 2 x$, so for any value $-\frac{1}{2} \leqslant t \leqslant \frac{1}{2}$, there is a suitable $x$. Furthermore, $A$ is expressed in terms of $t$ up to a sign, and both possible values of $A$ are achieved for the given $t$: to change the sign of $A$, it is sufficient to increase or decrease $x$ by $\pi$.