Simplify this trigonometric equation (identities) Please simplify these trigonometric identities and describe step by step (wording).
$\frac{\sec \theta-\csc \theta}{1-\cot\theta}$
 A: Let $A = \frac{\sec\theta - \csc\theta}{1-\cot\theta}$
$A = \frac{\frac{1}{\cos\theta}-\frac{1}{\sin\theta}}{1-\frac{\cos\theta}{\sin\theta}}$
Now we make common denominators like so:
$A = \frac{\frac{\sin\theta}{\sin\theta \cdot \cos\theta}-\frac{\cos\theta}{sin\theta\cdot \cos\theta}}{\frac{\sin\theta}{\sin\theta}-\frac{\cos\theta}{\sin\theta}}$
$A = \frac{\sin\theta - \cos\theta}{\sin\theta\cdot\cos\theta} \cdot \frac{\sin\theta}{\sin\theta-\cos\theta}$
Now we can cancel out leaving $A = \frac{1}{\cos\theta} = \sec\theta$
A: Change secϴ to 1/cosϴ, cscϴ to 1/sinϴ, and cotϴ to cosϴ/sinϴ so you have
$$\frac{\frac{1}{cosϴ}-\frac{1}{sinϴ}}{1-\frac{cosϴ}{sinϴ}}$$
Now multiply both the numerator and denominator of 1/cosϴ by sinϴ, the numerator and denominator of 1/sinϴ by cosϴ, and the numerator and denominator of 1 by sinϴ in order to add the two terms together (remember for fractions, if the denominators are equal, then just add the numerators together) so you have
$$\frac{\frac{sinϴ}{sinϴcosϴ}-\frac{cosϴ}{sinϴcosϴ}}{\frac{sinϴ}{sinϴ}-\frac{cosϴ}{sinϴ}}$$
Now you just add the numerators so you have
$$\frac{\frac{sinϴ-cosϴ}{sinϴcosϴ}}{\frac{sinϴ-cosϴ}{sinϴ}}$$
Now dividing by a fraction is the same as multiplying by the reciprocal of it so the operation turns into
$$\frac{sinϴ-cosϴ}{sinϴcosϴ}*\frac{sinϴ}{sinϴ-cosϴ}$$
Now you see that the numerator of the first term cancels with the denominator of the second term so you are left with
$$\frac{sinϴ}{sinϴcosϴ}$$
Now the sinϴ in the numerator cancels with the sinϴ in the denominator so you are left with
$$\frac{1}{cosϴ}$$
Or simply
$$secϴ$$
