Join IDNLearn.com and start exploring the answers to your most pressing questions. Get thorough and trustworthy answers to your queries from our extensive network of knowledgeable professionals.
Sagot :
To prove the given trigonometric identity:
[tex]\[ \frac{\sec \alpha}{1 + \tan \alpha} = \frac{\operatorname{cosec} \alpha}{1 + \cot \alpha} \][/tex]
we'll start by simplifying both sides of the equation separately and then show that they are equal.
### Left Side of the Identity
Consider the left side of the identity:
[tex]\[ \frac{\sec \alpha}{1 + \tan \alpha} \][/tex]
The other trigonometric identities we can use here are:
[tex]\[ \sec \alpha = \frac{1}{\cos \alpha} \quad \text{and} \quad \tan \alpha = \frac{\sin \alpha}{\cos \alpha} \][/tex]
Thus, we can rewrite the expression as:
[tex]\[ \frac{\frac{1}{\cos \alpha}}{1 + \frac{\sin \alpha}{\cos \alpha}} \][/tex]
Simplify the denominator:
[tex]\[ 1 + \frac{\sin \alpha}{\cos \alpha} = \frac{\cos \alpha + \sin \alpha}{\cos \alpha} \][/tex]
Therefore, the left side becomes:
[tex]\[ \frac{\frac{1}{\cos \alpha}}{\frac{\cos \alpha + \sin \alpha}{\cos \alpha}} = \frac{1}{\cos \alpha + \sin \alpha} \][/tex]
### Right Side of the Identity
Now consider the right side of the identity:
[tex]\[ \frac{\operatorname{cosec} \alpha}{1 + \cot \alpha} \][/tex]
Again, using the trigonometric identities:
[tex]\[ \operatorname{cosec} \alpha = \frac{1}{\sin \alpha} \quad \text{and} \quad \cot \alpha = \frac{\cos \alpha}{\sin \alpha} \][/tex]
Rewrite the expression as:
[tex]\[ \frac{\frac{1}{\sin \alpha}}{1 + \frac{\cos \alpha}{\sin \alpha}} \][/tex]
Simplify the denominator:
[tex]\[ 1 + \frac{\cos \alpha}{\sin \alpha} = \frac{\sin \alpha + \cos \alpha}{\sin \alpha} \][/tex]
Therefore, the right side becomes:
[tex]\[ \frac{\frac{1}{\sin \alpha}}{\frac{\sin \alpha + \cos \alpha}{\sin \alpha}} = \frac{1}{\sin \alpha + \cos \alpha} \][/tex]
### Conclusion
Now, we observe that both simplified expressions for the left and right sides are indeed equal:
[tex]\[ \frac{1}{\cos \alpha + \sin \alpha} = \frac{1}{\sin \alpha + \cos \alpha} \][/tex]
Consequently, we have:
[tex]\[ \frac{\sec \alpha}{1 + \tan \alpha} = \frac{\operatorname{cosec} \alpha}{1 + \cot \alpha} \][/tex]
Therefore, the identity [tex]\(\frac{\sec \alpha}{1 + \tan \alpha} = \frac{\operatorname{cosec} \alpha}{1 + \cot \alpha}\)[/tex] is proven to be true.
[tex]\[ \frac{\sec \alpha}{1 + \tan \alpha} = \frac{\operatorname{cosec} \alpha}{1 + \cot \alpha} \][/tex]
we'll start by simplifying both sides of the equation separately and then show that they are equal.
### Left Side of the Identity
Consider the left side of the identity:
[tex]\[ \frac{\sec \alpha}{1 + \tan \alpha} \][/tex]
The other trigonometric identities we can use here are:
[tex]\[ \sec \alpha = \frac{1}{\cos \alpha} \quad \text{and} \quad \tan \alpha = \frac{\sin \alpha}{\cos \alpha} \][/tex]
Thus, we can rewrite the expression as:
[tex]\[ \frac{\frac{1}{\cos \alpha}}{1 + \frac{\sin \alpha}{\cos \alpha}} \][/tex]
Simplify the denominator:
[tex]\[ 1 + \frac{\sin \alpha}{\cos \alpha} = \frac{\cos \alpha + \sin \alpha}{\cos \alpha} \][/tex]
Therefore, the left side becomes:
[tex]\[ \frac{\frac{1}{\cos \alpha}}{\frac{\cos \alpha + \sin \alpha}{\cos \alpha}} = \frac{1}{\cos \alpha + \sin \alpha} \][/tex]
### Right Side of the Identity
Now consider the right side of the identity:
[tex]\[ \frac{\operatorname{cosec} \alpha}{1 + \cot \alpha} \][/tex]
Again, using the trigonometric identities:
[tex]\[ \operatorname{cosec} \alpha = \frac{1}{\sin \alpha} \quad \text{and} \quad \cot \alpha = \frac{\cos \alpha}{\sin \alpha} \][/tex]
Rewrite the expression as:
[tex]\[ \frac{\frac{1}{\sin \alpha}}{1 + \frac{\cos \alpha}{\sin \alpha}} \][/tex]
Simplify the denominator:
[tex]\[ 1 + \frac{\cos \alpha}{\sin \alpha} = \frac{\sin \alpha + \cos \alpha}{\sin \alpha} \][/tex]
Therefore, the right side becomes:
[tex]\[ \frac{\frac{1}{\sin \alpha}}{\frac{\sin \alpha + \cos \alpha}{\sin \alpha}} = \frac{1}{\sin \alpha + \cos \alpha} \][/tex]
### Conclusion
Now, we observe that both simplified expressions for the left and right sides are indeed equal:
[tex]\[ \frac{1}{\cos \alpha + \sin \alpha} = \frac{1}{\sin \alpha + \cos \alpha} \][/tex]
Consequently, we have:
[tex]\[ \frac{\sec \alpha}{1 + \tan \alpha} = \frac{\operatorname{cosec} \alpha}{1 + \cot \alpha} \][/tex]
Therefore, the identity [tex]\(\frac{\sec \alpha}{1 + \tan \alpha} = \frac{\operatorname{cosec} \alpha}{1 + \cot \alpha}\)[/tex] is proven to be true.
Your participation means a lot to us. Keep sharing information and solutions. This community grows thanks to the amazing contributions from members like you. For trustworthy and accurate answers, visit IDNLearn.com. Thanks for stopping by, and see you next time for more solutions.
