Question

Find set of all possible values of a in [-pi,pi]such that sqrt((1-sina)/(1+sina)) ls equal to (seca – tana)

Answer 1

Find set of all possible values of $a$ in $[-pi, pi]$ such that $\sqrt{1 + \sin a} / (1 + \sin a)$ is equal to $(\sec a - \tan a)$.

Solution.

1. Rewrite the equation in the form

$$\sqrt{\frac{1 - \sin a}{1 + \sin a}} = \frac{1 - \sin a}{\cos a}$$

2. Describe the set of allowed values of $a$.

$$\frac{1 - \sin a}{1 + \sin a} \geq 0,$$ $$1 + \sin a \neq 0,$$ $$\frac{1 - \sin a}{\cos a} \geq 0,$$ $$\cos a \neq 0.$$

3. Solve the equation (1).

Squaring both parts

$$\frac{1 - \sin a}{1 + \sin a} = \frac{(1 - \sin a)^2}{(\cos a)^2},$$

taking out common multiple

$$(1 - \sin a) \left(\frac{1 - \sin a}{\cos^2 a} - \frac{1}{1 + \sin a}\right) = 0,$$

and reducing to the same denominator

$$(1 - \sin a) \left(\frac{1 - \sin^2 a - \cos^2 a}{\cos^2 a (1 + \sin a)}\right) = 0$$

we have

$$(1 - \sin a) * 0 = 0.$$

So, the values of $a$ satisfying requirements (2)-(5) are solutions of the equation (1).

4. Solve (2)-(5).

We have

$$\sin a \neq -1 \Rightarrow a \neq -\frac{\pi}{2} + 2\pi k, k \in \mathbb{Z},$$ $$\cos a \neq 0 \Rightarrow a \neq \pm \frac{\pi}{2} + 2\pi k,$$ $$\cos a > 0 \Rightarrow a \in \left(-\frac{\pi}{2} + 2\pi k, \frac{\pi}{2} + 2\pi k\right).$$

So,

$$a \in \left(-\frac{\pi}{2} + 2\pi k, \frac{\pi}{2} + 2\pi k\right).$$

5. Finally taking in mind that $a \in [-\pi, \pi]$ we get $k = 0$ and $a \in (-\frac{\pi}{2}, \frac{\pi}{2})$.

Answer: $a \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$.