Perform the indicated operations and simplify the result so there are no quotients.
-$$\frac { \sin \theta } { 1 + \sin \theta } - \frac { \sin \theta } { 1 - \sin \theta }$$
A) $1 + \cot \theta$
B) $\sec \theta \csc \theta$
C) $\sin \theta \tan \theta$
D) $- 2 \tan ^ { 2 } \theta$

Question

Quizplus · Accepted Answer

To simplify the given expression, we first find a common denominator and combine the fractions: $$\frac{\sin \theta (1 - \sin \theta) - \sin \theta (1 + \sin \theta)}{(1 + \sin \theta)(1 - \sin \theta)}$$ Simplifying the numerator and denominator gives: $$\frac{\sin \theta - \sin^2 \theta - \sin \theta - \sin^2 \theta}{1 - \sin^2 \theta}$$ Further simplification of the numerator and using the identity $1 - \sin^2 \theta = \cos^2 \theta$ in the denominator gives: $$\frac{-2\sin^2 \theta}{\cos^2 \theta}$$ Recognizing that $\sin^2 \theta / \cos^2 \theta = \tan^2 \theta$, we get: $$-2\tan^2 \theta$$

Perform the Indicated Operations and Simplify the Result So There sin⁡θ1+sin⁡θ−sin⁡θ1−sin⁡θ\frac { \sin \theta } { 1 + \sin \theta } - \frac { \sin \theta } { 1 - \sin \theta }1+sinθsinθ​−1−sinθsinθ​

Perform the Indicated Operations and Simplify the Result So There $\frac { \sin \theta } { 1 + \sin \theta } - \frac { \sin \theta } { 1 - \sin \theta }$