Answer:
Proof below
Step-by-step explanation:
Trigonometric Identities
We'll prove that:
[tex]\displaystyle \frac{\sin A+\cos A}{\sec A+\csc A}=\sin A*\cos A[/tex]
Recall:
[tex]\displaystyle \sec A =\frac{1}{\cos A}[/tex]
[tex]\displaystyle \csc A =\frac{1}{\sin A}[/tex]
Applying those definitions:
[tex]\displaystyle \frac{\sin A+\cos A}{\sec A+\csc A}=\frac{\sin A+\cos A}{\frac{1}{\cos A}+\frac{1}{\sin A}}[/tex]
Adding the fractions:
[tex]\displaystyle \frac{\sin A+\cos A}{\sec A+\csc A}=\frac{\sin A+\cos A}{\frac{\sin A+\cos A}{\cos A*\sin A}}[/tex]
Dividing the fractions:
[tex]\displaystyle \frac{\sin A+\cos A}{\sec A+\csc A}=(\sin A+\cos A)*\frac{\cos A*\sin A}{\sin A+\cos A}[/tex]
Simplifying:
[tex]\displaystyle \frac{\sin A+\cos A}{\sec A+\csc A}=\cos A*\sin A[/tex]
Hence proved
