Answer :

To prove the trigonometric identity [tex]\(\frac{1}{\cot \theta + \tan \theta} = \sin \theta \cos \theta\)[/tex], we will proceed with a step-by-step simplification of the left-hand side.

1. Express [tex]\(\cot \theta\)[/tex] in terms of [tex]\(\tan \theta\)[/tex]:
- Recall that [tex]\(\cot \theta = \frac{1}{\tan \theta}\)[/tex].
- Therefore, [tex]\(\cot \theta + \tan \theta = \frac{1}{\tan \theta} + \tan \theta\)[/tex].

2. Combine the terms on the left-hand side with a common denominator:
- Using a common denominator, we have:
[tex]\[ \frac{1}{\cot \theta + \tan \theta} = \frac{1}{\frac{1}{\tan \theta} + \tan \theta} = \frac{1}{\frac{1 + \tan^2 \theta}{\tan \theta}}. \][/tex]

3. Simplify the expression inside the denominator:
- The expression [tex]\(\frac{1 + \tan^2 \theta}{\tan \theta}\)[/tex] in the denominator can be simplified using the Pythagorean identity [tex]\(1 + \tan^2 \theta = \sec^2 \theta\)[/tex]:
[tex]\[ \frac{1 + \tan^2 \theta}{\tan \theta} = \frac{\sec^2 \theta}{\tan \theta}. \][/tex]

4. Rewrite [tex]\(\sec^2 \theta\)[/tex] and [tex]\(\tan \theta\)[/tex] in terms of [tex]\(\sin \theta\)[/tex] and [tex]\(\cos \theta\)[/tex]:
- Recall that [tex]\(\sec \theta = \frac{1}{\cos \theta}\)[/tex] and [tex]\(\tan \theta = \frac{\sin \theta}{\cos \theta}\)[/tex]. Therefore, [tex]\(\sec^2 \theta = \frac{1}{\cos^2 \theta}\)[/tex] and [tex]\(\tan \theta = \frac{\sin \theta}{\cos \theta}\)[/tex].
- Substitute these identities into the expression:
[tex]\[ \frac{\sec^2 \theta}{\tan \theta} = \frac{\frac{1}{\cos^2 \theta}}{\frac{\sin \theta}{\cos \theta}} = \frac{1}{\cos^2 \theta} \cdot \frac{\cos \theta}{\sin \theta} = \frac{1}{\cos \theta \sin \theta}. \][/tex]

5. Take the reciprocal of the denominator to complete the simplification:
[tex]\[ \frac{1}{\frac{1}{\cos \theta \sin \theta}} = \cos \theta \sin \theta. \][/tex]

So, we have shown that:
[tex]\[ \frac{1}{\cot \theta + \tan \theta} = \cos \theta \sin \theta. \][/tex]

Therefore, the identity [tex]\(\frac{1}{\cot \theta + \tan \theta} = \sin \theta \cos \theta\)[/tex] is proven.