To prove the given equation [tex]\(\frac{\cos \theta}{\operatorname{cosec} \theta + 1} + \frac{\cos \theta}{\operatorname{cosec} \theta - 1} = 2 \tan \theta\)[/tex], let's go through the steps systematically.
1. Express [tex]\(\operatorname{cosec} \theta\)[/tex]:
[tex]\[
\operatorname{cosec} \theta = \frac{1}{\sin \theta}
\][/tex]
2. Substitute [tex]\(\operatorname{cosec} \theta\)[/tex] into the given equation:
[tex]\[
\frac{\cos \theta}{\frac{1}{\sin \theta} + 1} + \frac{\cos \theta}{\frac{1}{\sin \theta} - 1}
\][/tex]
3. Simplify the denominators:
[tex]\[
\frac{\cos \theta}{\frac{1 + \sin \theta}{\sin \theta}} + \frac{\cos \theta}{\frac{1 - \sin \theta}{\sin \theta}}
\][/tex]
Which simplifies to:
[tex]\[
\frac{\cos \theta \sin \theta}{1 + \sin \theta} + \frac{\cos \theta \sin \theta}{1 - \sin \theta}
\][/tex]
4. Combine the fractions:
[tex]\[
\cos \theta \sin \theta \left(\frac{1}{1 + \sin \theta} + \frac{1}{1 - \sin \theta}\right)
\][/tex]
5. Find common denominator:
[tex]\[
\cos \theta \sin \theta \left(\frac{1 - \sin \theta + 1 + \sin \theta}{(1 + \sin \theta)(1 - \sin \theta)}\right)
\][/tex]
Notice that:
[tex]\[
(1 + \sin \theta)(1 - \sin \theta) = 1 - \sin^2 \theta = \cos^2 \theta
\][/tex]
So:
[tex]\[
\cos \theta \sin \theta \left(\frac{2}{\cos^2 \theta}\right)
\][/tex]
6. Simplify the expression:
[tex]\[
\frac{2 \cos \theta \sin \theta}{\cos^2 \theta} = 2 \left(\frac{\sin \theta}{\cos \theta}\right) = 2 \tan \theta
\][/tex]
Therefore, we have shown that:
[tex]\[
\frac{\cos \theta}{\operatorname{cosec} \theta + 1} + \frac{\cos \theta}{\operatorname{cosec} \theta - 1} = 2 \tan \theta
\][/tex]
This completes the proof.
