(cosθ)(tanθ - secθ) = sinθ - 1

Understand the Problem

The question is asking to solve the equation involving trigonometric functions: (cosθ)(tanθ - secθ) = sinθ - 1. We will need to manipulate the trigonometric identities to simplify or solve for variable θ.

Answer

The equation is satisfied for all $\theta$, except where $\cos \theta = 0$.

Steps to Solve

Rewrite the equation using identities

Start by rewriting the trigonometric functions in terms of sine and cosine.

$$ \cos \theta (\tan \theta - \sec \theta) = \sin \theta - 1 $$

Recall that:

$\tan \theta = \frac{\sin \theta}{\cos \theta}$
$\sec \theta = \frac{1}{\cos \theta}$

Substituting these into the equation gives:

$$ \cos \theta \left(\frac{\sin \theta}{\cos \theta} - \frac{1}{\cos \theta}\right) = \sin \theta - 1 $$

Simplify the left side

Now, simplify the left side of the equation:

$$ \cos \theta \left(\frac{\sin \theta - 1}{\cos \theta}\right) = \sin \theta - 1 $$

The $\cos \theta$ cancels:

$$ \sin \theta - 1 = \sin \theta - 1 $$

Identify the solution

Since both sides of the equation are equal, it is true for all $\theta$ where the functions are defined, except for points where $\cos \theta = 0$ (e.g., $\theta = \frac{\pi}{2} + k\pi$, where $k$ is any integer).

The equation is satisfied for all $\theta$, except where $\cos \theta = 0$.

More Information

This means that the equation holds true for almost every angle $\theta$, reflecting the periodic nature of trigonometric functions.

Tips

Not recognizing that both sides simplify to the same expression.
Forgetting to consider the points where $\cos \theta = 0$, which are excluded from the solution.

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