To prove the given equation:

$\sqrt{\frac{\sec A - 1}{\sec A + 1}} + \sqrt{\frac{\sec A + 1}{\sec A - 1}} = 2\csc A$

Step-by-Step Proof

1. Simplify Each Square Root: Let’s consider the terms under the square roots.

   • Rewrite $\sec A$ in terms of $\cos A$: $\sec A = \frac{1}{\cos A}, \quad \therefore \sec A - 1 = \frac{1 - \cos A}{\cos A}, \quad \sec A + 1 = \frac{1 + \cos A}{\cos A}.$

   Substituting into the expressions under the square roots:

   $\sqrt{\frac{\sec A - 1}{\sec A + 1}} = \sqrt{\frac{\frac{1 - \cos A}{\cos A}}{\frac{1 + \cos A}{\cos A}}} = \sqrt{\frac{1 - \cos A}{1 + \cos A}}.$

   Similarly:

   $\sqrt{\frac{\sec A + 1}{\sec A - 1}} = \sqrt{\frac{1 + \cos A}{1 - \cos A}}.$

2. Combine the Two Terms: Substitute the simplified forms back into the equation: $\sqrt{\frac{1 - \cos A}{1 + \cos A}} + \sqrt{\frac{1 + \cos A}{1 - \cos A}}.$

   Multiply and divide by $\sqrt{(1 - \cos A)(1 + \cos A)}$ (which simplifies to $\sin^2 A$):

   $\sqrt{\frac{1 - \cos A}{1 + \cos A}} + \sqrt{\frac{1 + \cos A}{1 - \cos A}} = \frac{\sqrt{(1 - \cos A)(1 + \cos A)} + \sqrt{(1 + \cos A)(1 - \cos A)}}{\sin A}.$

   Using the identity $1 - \cos^2 A = \sin^2 A$, the numerator simplifies to:

   $\frac{\sin A + \sin A}{\sin A} = 2.$

3. Express in Terms of $\csc A$: Since $\csc A = \frac{1}{\sin A}$, the equation simplifies to: $2 \csc A.$

Thus, the left-hand side equals the right-hand side, proving the equation: $\sqrt{\frac{\sec A - 1}{\sec A + 1}} + \sqrt{\frac{\sec A + 1}{\sec A - 1}} = 2 \csc A.$

Would you like a deeper explanation for any step? Here are 5 related questions for further exploration:

1. How is the identity $\sec^2 A - 1 = \tan^2 A$ derived from trigonometric principles?
2. Why does $\sqrt{\frac{1 - \cos A}{1 + \cos A}}$ simplify to $\sin A$ using trigonometric identities?
3. Can this proof be done graphically or using numerical methods for specific values of $A$?
4. What are some real-world applications of proving trigonometric identities like this one?
5. How does this equation relate to other trigonometric equations involving $\csc A$ and $\sec A$?

Tip: Practice rewriting trigonometric identities in terms of sine and cosine—it simplifies proofs significantly!