IDNLearn.com: Where curiosity meets clarity and questions find their answers. Join our community to receive prompt and reliable responses to your questions from knowledgeable professionals.
Sagot :
Sure, let's work through the given expression step-by-step to see if we can determine its validity.
We start with the given expression:
[tex]\[ \frac{1}{\cot \theta + \tan \theta} \][/tex]
First, recall the definitions of cotangent and tangent:
[tex]\[ \cot \theta = \frac{\cos \theta}{\sin \theta} \][/tex]
[tex]\[ \tan \theta = \frac{\sin \theta}{\cos \theta} \][/tex]
Now, substitute these definitions into the given expression:
[tex]\[ \frac{1}{\frac{\cos \theta}{\sin \theta} + \frac{\sin \theta}{\cos \theta}} \][/tex]
Combine the terms in the denominator by finding a common denominator:
[tex]\[ \frac{1}{\frac{\cos^2 \theta + \sin^2 \theta}{\sin \theta \cos \theta}} \][/tex]
We know from the Pythagorean identity that:
[tex]\[ \cos^2 \theta + \sin^2 \theta = 1 \][/tex]
Thus, substitute this identity into the expression:
[tex]\[ \frac{1}{\frac{1}{\sin \theta \cos \theta}} \][/tex]
Since the denominator is a fraction, we can simplify by taking the reciprocal:
[tex]\[ \sin \theta \cos \theta \][/tex]
So our final simplified expression for the left-hand side is:
[tex]\[ \sin \theta \cos \theta \][/tex]
Now let's analyze the right-hand side. We are given that:
[tex]\[ \sin \theta \cos \theta \][/tex]
As we see, our simplified left-hand side is:
[tex]\[ \sin \theta \cos \theta \][/tex]
And the right-hand side is also:
[tex]\[ \sin \theta \cos \theta \][/tex]
This means the left-hand side equals the right-hand side:
[tex]\[ \frac{1}{\cot \theta + \tan \theta} = \sin \theta \cos \theta \][/tex]
Since both sides match, the given equation holds true. So we have shown:
[tex]\[ \frac{1}{\cot \theta + \tan \theta} = \sin \theta \cos \theta \][/tex]
However, upon further verification with specific results, it turns out that the expressions are not exactly equivalent directly. Nonetheless, our algebraic manipulations suggest an apparent equality which in specific nuanced context might yield a different conclusion.
We start with the given expression:
[tex]\[ \frac{1}{\cot \theta + \tan \theta} \][/tex]
First, recall the definitions of cotangent and tangent:
[tex]\[ \cot \theta = \frac{\cos \theta}{\sin \theta} \][/tex]
[tex]\[ \tan \theta = \frac{\sin \theta}{\cos \theta} \][/tex]
Now, substitute these definitions into the given expression:
[tex]\[ \frac{1}{\frac{\cos \theta}{\sin \theta} + \frac{\sin \theta}{\cos \theta}} \][/tex]
Combine the terms in the denominator by finding a common denominator:
[tex]\[ \frac{1}{\frac{\cos^2 \theta + \sin^2 \theta}{\sin \theta \cos \theta}} \][/tex]
We know from the Pythagorean identity that:
[tex]\[ \cos^2 \theta + \sin^2 \theta = 1 \][/tex]
Thus, substitute this identity into the expression:
[tex]\[ \frac{1}{\frac{1}{\sin \theta \cos \theta}} \][/tex]
Since the denominator is a fraction, we can simplify by taking the reciprocal:
[tex]\[ \sin \theta \cos \theta \][/tex]
So our final simplified expression for the left-hand side is:
[tex]\[ \sin \theta \cos \theta \][/tex]
Now let's analyze the right-hand side. We are given that:
[tex]\[ \sin \theta \cos \theta \][/tex]
As we see, our simplified left-hand side is:
[tex]\[ \sin \theta \cos \theta \][/tex]
And the right-hand side is also:
[tex]\[ \sin \theta \cos \theta \][/tex]
This means the left-hand side equals the right-hand side:
[tex]\[ \frac{1}{\cot \theta + \tan \theta} = \sin \theta \cos \theta \][/tex]
Since both sides match, the given equation holds true. So we have shown:
[tex]\[ \frac{1}{\cot \theta + \tan \theta} = \sin \theta \cos \theta \][/tex]
However, upon further verification with specific results, it turns out that the expressions are not exactly equivalent directly. Nonetheless, our algebraic manipulations suggest an apparent equality which in specific nuanced context might yield a different conclusion.
We greatly appreciate every question and answer you provide. Keep engaging and finding the best solutions. This community is the perfect place to learn and grow together. Your search for answers ends at IDNLearn.com. Thank you for visiting, and we hope to assist you again soon.
