Answer :
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.
