Question

(sinA + cosA)/ (secA + cosecA) = sinA* cosA​

Answer 1

Answer:

Proof below

Step-by-step explanation:

Trigonometric Identities

We'll prove that:

$\displaystyle \frac{\sin A+\cos A}{\sec A+\csc A}=\sin A*\cos A$

Recall:

$\displaystyle \sec A =\frac{1}{\cos A}$

$\displaystyle \csc A =\frac{1}{\sin A}$

Applying those definitions:

$\displaystyle \frac{\sin A+\cos A}{\sec A+\csc A}=\frac{\sin A+\cos A}{\frac{1}{\cos A}+\frac{1}{\sin A}}$

Adding the fractions:

$\displaystyle \frac{\sin A+\cos A}{\sec A+\csc A}=\frac{\sin A+\cos A}{\frac{\sin A+\cos A}{\cos A*\sin A}}$

Dividing the fractions:

$\displaystyle \frac{\sin A+\cos A}{\sec A+\csc A}=(\sin A+\cos A)*\frac{\cos A*\sin A}{\sin A+\cos A}$

Simplifying:

$\displaystyle \frac{\sin A+\cos A}{\sec A+\csc A}=\cos A*\sin A$

Hence proved