prove Identity $$ \frac{1-\cos x}{\sin x}-\frac{\sin x}{1+\cos x}=0 $$

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UpStudy AI · Accepted Answer

1. Start with the given expression:
   $$
   \frac{1-\cos x}{\sin x} - \frac{\sin x}{1+\cos x}
   $$

2. Consider rewriting the second fraction by rationalizing its denominator. Multiply the numerator and denominator of the second term by $$(1-\cos x)$$:
   $$
   \frac{\sin x}{1+\cos x} = \frac{\sin x(1-\cos x)}{(1+\cos x)(1-\cos x)}
   $$

3. Notice that the denominator simplifies using the difference of squares:
   $$
   (1+\cos x)(1-\cos x) = 1 - \cos^2 x
   $$

4. Recall the Pythagorean identity:
   $$
   1 - \cos^2 x = \sin^2 x
   $$
   Therefore,
   $$
   \frac{\sin x(1-\cos x)}{(1+\cos x)(1-\cos x)} = \frac{\sin x(1-\cos x)}{\sin^2 x}
   $$

5. Simplify the fraction by canceling a common factor of $$\sin x$$ in the numerator and denominator:
   $$
   \frac{\sin x(1-\cos x)}{\sin^2 x} = \frac{1-\cos x}{\sin x}
   $$

6. Substitute this back into the original expression:
   $$
   \frac{1-\cos x}{\sin x} - \frac{1-\cos x}{\sin x} = 0
   $$

7. Thus, the identity is proven:
   $$
   \frac{1-\cos x}{\sin x} - \frac{\sin x}{1+\cos x} = 0
   $$
The identity $$ \frac{1-\cos x}{\sin x} - \frac{\sin x}{1+\cos x} = 0 $$ is proven by simplifying both terms and showing they cancel each other out.

prove Identity \( \frac{1-\cos x}{\sin x}-\frac{\sin x}{1+\cos x}=0 \)

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