Question

Prove that 1/(1+cosx) = csc^2(x) - cotx/sinx

I've tried a few things and can't seem to get this right... I think the first step is to change csc^2x into (1+cot^2x) and then get a common denominator of sinx, but either this is the wrong way or I'm making more mistakes further on... any ideas? Thanks.

Answer 1

well your idea is absolutely right actually

\[ \large \frac{1}{\sin^2(x)}- \frac{\cot x \sin x}{\sin ^2x}=\frac{1-\cos(x)}{1-\cos^2x } \\
\frac{1-\cos(x)}{1-\cos^2x} =\frac{1-\cos x}{(1+\cos x)(1-\cos x) }= \frac{1}{1+\cos(x)}\]

Answer 2

0. Given
\(\large\frac{1}{1+\cos x} = \csc ^2x - \frac{\cot x}{\sin x}\)

1. Replace csc^2x with 1/sin^2x:
\(\large\frac{1}{1+\cos x} = \frac{1}{\sin^2x} - \frac{\cot x}{\sin x}\)

2. Rewrite cotx/sinx as 1/sinx times cos x/sin x:
\(\large\frac{1}{1+\cos x} = \frac{1}{\sin^2x} - \frac{1}{\sin x} \dot\ \frac{\cos x}{\sin x}\)

3. Multiply:
\(\large\frac{1}{1+\cos x} = \frac{1}{\sin^2x} - \frac{\cos x}{\sin^2x}\)

4. Combine fractions:
\(\large\frac{1}{1+\cos x} = \frac{1-cos x}{\sin^2x}\)

5. Cross Multiply:
\(\sin^2x = (1+\cos x)(1-\cos x)\)

6. Muliply the right side to get:
\(\sin^2x = 1 - \cos^2 x\)

7. Add cos^2x to both sides to get:
\(\sin^2x + cos^2x = 1\)

8. Simplify:
\(1 = 1\)

9. Solution:
True

Answer 3

Thank you both! I ended up getting a common denominator of sin^2x(1+cosx), converting the numerator to sin^2x and cancelling it out to get 1/(1+cosx).