This is not true.  This is because when squaring both sides and then plugging back into the original equations, some of our solutions may be extraneous solutions.  Therefore, when solving a trigonometric equation by squaring both sides, all solutions found must be plugged back into the original equation and validated.

We should only square by both sides when all other identities are not able to be used in an equation.  Quite often, you will find that a trigonometric identity can be used to simplify an equation.  Squaring both sides is ultimately trying to produce a trigonometric identity in order to solve for the equation.

We begin with our original equation:

$\displaystyle 1 + cot(\theta) = csc(\theta)$

$\displaystyle (1+cot(\theta))^2 = (csc(\theta))^2$

$\displaystyle 1 + 2cot(\theta) + cot^2(\theta) = csc^2(\theta)$

$\displaystyle 1 + 2cot(\theta) + cot^2(\theta) = 1 + cot^2(\theta)$                 (Pythagorean Identity)

$\displaystyle 2cot(\theta) = 0$

$\displaystyle cot(\theta) = 0$

Looking at the unit circle we see that $\displaystyle cot(\theta) = 0$ at $\displaystyle \frac {\pi}{2}$ and $\displaystyle \frac{3\pi}{2}$.  We must plug these back into our original equation to validate them.

Checking $\displaystyle \theta = \frac{\pi}{2}$

$\displaystyle 1 + cot(\frac{\pi}{2}) = csc(\frac{\pi}{2})$

$\displaystyle 1 + 0 = 1$

$\displaystyle 1 = 1$

Checking $\displaystyle \theta = \frac{3\pi}{2}$

$\displaystyle 1 + cot(\frac{3\pi}{2}) = \frac{3\pi}{2}$

$\displaystyle 1 + 0 = -1$

$\displaystyle 1 \neq -1$

And so our only solution is $\displaystyle \theta = \frac{\pi}{2}$

We begin with our original equation:

$\displaystyle -\sqrt{3} + tan(\theta) = sec(\theta)$

$\displaystyle (-\sqrt{3} + tan(\theta))^2$$\displaystyle = (sec(\theta))^2 (sec(\theta))^2sec(\theta))^2c(\theta))^2(\theta))^2(\theta))^2(\tan))^2))^2^2$

$\displaystyle 3 - 2\sqrt{3}tan(\theta) + tan^2(\theta) = sec^2(\theta)$

$\displaystyle 3 - 2\sqrt{3}tan(\theta) + tan^2(\theta) = 1 + tan^2(\theta)$               (Pythagorean Identity)

$\displaystyle -2\sqrt{3}tan(\theta) = -2$

$\displaystyle tan(\theta) = \frac{1}{\sqrt{3}}$

From the unit circle, we see that $\displaystyle tan(\theta) = \frac{1}{\sqrt{3}}$$ when $$\theta = \frac{\pi}{6}, \frac{7\pi}{6}$. We must check both of these solutions in the original equation.

Checking $\displaystyle \theta = \frac{\pi}{6}$

$\displaystyle -\sqrt{3} + tan(\frac{\pi}{6}) = sec(\frac{\pi}{6})$

$\displaystyle -\sqrt{3} + \frac{\sqrt{3}}{3} = \frac{2\sqrt{3}}{3}$

$\displaystyle \frac{\sqrt{3}}{3} \neq \frac{5\sqrt{3}}{3}$

Checking  $\displaystyle \theta = \frac{7\pi}{6}$

$\displaystyle -\sqrt{3} + tan(\frac{7\pi}{6}) = sec(\frac{7\pi}{6})$

$\displaystyle -\sqrt{3} + \frac{\sqrt{3}}{3} = -\frac{2\sqrt{3}}{3}$

$\displaystyle \frac{\sqrt{3}}{3} = \frac{\sqrt{3}}{3}$

So we see our only solution is $\displaystyle \theta = \frac{\sqrt{\pi}}{6}$

We begin with our original equation

$\displaystyle tan(\theta) + 1 = sec(\theta)$

$\displaystyle (tan(\theta) + 1)^2 = (sec(\theta))^2$

$\displaystyle tan^2(\theta) + 2tan(\theta) + 1 = sec^2(\theta)$

$\displaystyle tan^2(\theta) + 2tan(\theta) + 1 = tan^2(\theta) + 1$                                  (Pythagorean Identity)

$\displaystyle 2tan(\theta) = 0$

$\displaystyle tan(\theta) = 0$

$\displaystyle \frac{sin(\theta)}{cos(\theta)} = 0$                                                                                            (substitution)

Using this form, we see we really only need to consider when $\displaystyle sin(\theta) = 0$  at $\displaystyle 0$  and $\displaystyle \pi$.  Now we must plug these values into the original equation to check and see if they are both acceptable solutions to our problem.

Checking $\displaystyle \theta = 0$:

$\displaystyle tan(0) + 1 = sec(0)$

$\displaystyle 0 + 1 = 1$

$\displaystyle 1 = 1$

Checking $\displaystyle \theta = \pi$

$\displaystyle tan(\pi) + 1 = sec(\pi)$

$\displaystyle 0 + 1 = -1$

$\displaystyle 1 \neq -1$

By checking our solutions we see the only solution to our equation is $\displaystyle \theta = 0$.

We begin with our original equation.

$\displaystyle cos(\theta) + sin(\theta) = 1$

$\displaystyle (cos(\theta) + sin(\theta))^2 = 1^2$

$\displaystyle cos^2(\theta) + 2cos(\theta)sin(theta) + sin^2(\theta) = 1$

$\displaystyle cos^2(\theta) + sin^2(\theta) + 2cos(\theta)sin(\theta) = 1$

$\displaystyle 1 + 2cos(\theta)sin(\theta) = 1$                                             (Pythagorean Identity)

$\displaystyle 2cos(\theta)sin(\theta) = 0$

$\displaystyle sin(2\theta) = 0$                                                                    (Double-Angle Formula)

We know that $\displaystyle sin(2\theta)$ will be equal to $\displaystyle 0$ for when $\displaystyle \theta$ is any multiple of and when $\displaystyle \theta = 0$.  We need to check both solutions (we will simply check $\displaystyle \pi$ for simplicity) to make sure they are valid solutions.

Checking $\displaystyle \theta = 0$:

$\displaystyle cos(0) + sin(0) = 1$

$\displaystyle 1 + 0 = 1$

$\displaystyle 1 = 1$

Checking $\displaystyle \theta = \pi$

$\displaystyle cos(\pi) + sin(\pi) = 1$

$\displaystyle -1 + 0 = 1$

$\displaystyle -1 \neq 1$

By checking our solutions, we see the only solution to this equation is $\displaystyle \theta = 0$

This one is not as straight-forward.  We must manipulate the original equation before squaring both sides.

$\displaystyle \sqrt{5}sin(\theta) + cos(\theta) = 1$

$\displaystyle \sqrt{5}sin(\theta) = 1 - cos(\theta)$

$\displaystyle (\sqrt{5}sin(\theta))^2 = (1 - cos(\theta))^2$

$\displaystyle 5sin^2(\theta) = 1 - 2cos(\theta) + cos^2(\theta)$

$\displaystyle 5sin^2(\theta) = 1 + cos^2(\theta) - 2cos(\theta)$

$\displaystyle 5(1 - cos^2(\theta) ) = 1 + cos^2(\theta) - 2cos(\theta)$                                     (Pythagorean Identity)

$\displaystyle 5 - 5cos^2(\theta) = 1 +cos^2(\theta) - 2cos(\theta)$

$\displaystyle 0 = 6cos^2(\theta) - 2cos(\theta) - 4$

$\displaystyle 0 = 3cos^2(\theta) - cos(\theta) - 2$                                                                 (divide both sides by 2)

$\displaystyle 0 = (3cos(\theta) + 2)(cos(\theta) -1)$

Solving for each:

$\displaystyle 3cos(\theta) + 2 = 0$

$\displaystyle 3cos(\theta) = -2$

$\displaystyle cos(\theta) = \frac{-2}{3}$

$\displaystyle \theta = cos^{-1}(\frac{-2}{3})$

$\displaystyle \theta = 2.3$   radians

Or 

$\displaystyle cos(\theta) - 1 = 0$

$\displaystyle cos(\theta) = 1$

From the unit circle we know that $\displaystyle cos(\theta) = 1$ when $\displaystyle \theta = 0, 2\pi$.

So now we must go back and check all of our solutions.

Checking  $\displaystyle \theta = 2.3$

$\displaystyle \sqrt{5}sin(2.3) + cos(2.3) = 1$

$\displaystyle \sqrt{5}(0.75) - 0.67 = 1$

$\displaystyle 1.67 - 0.67 = 1$

$\displaystyle 1 = 1$

Checking $\displaystyle \theta = 0$ (this is also equal to checking $\displaystyle \theta = 2\pi$)

$\displaystyle \sqrt{5}sin(0) + cos(0) = 1$

$\displaystyle 0 + 1 = 1$

$\displaystyle 1 = 1$

Both of our solutions are correct.

Our first line of defense when solving trigonometric functions is using a familiar identity/formula such as the Pythagorean Identities or the Double Angle Formulas.  When we are unable to use an identity or formula we are able to square both sides of the equation and with further manipulation we are usually able to produce one of these identities thus simplifying our problem and making it easier to solve.