If A+B=90° then show that (sinA+SecA)/sinA = 2tanAtanB

To demonstrate that \((\sin A + \sec A) / \sin A = 2 \tan A \tan B\) given that \(A + B = 90^\circ\), we can utilize some trigonometric identities and relationships. Let's delve into the proof step-by-step.

Step 1: Express Secant in Terms of Sine

Recall that the secant function is the reciprocal of cosine. Therefore, we can write:

\(\sec A = \frac{1}{\cos A}\)

Using the co-function identity, since \(A + B = 90^\circ\), we know that:

\(\cos A = \sin B\)

Thus, we can express secant as:

\(\sec A = \frac{1}{\sin B}\)

Step 2: Rewrite the Left Side of the Equation

Now, substituting \(\sec A\) in our original equation, we have:

\(\frac{\sin A + \sec A}{\sin A} = \frac{\sin A + \frac{1}{\sin B}}{\sin A}\)

This simplifies to:

\(\frac{\sin A}{\sin A} + \frac{1/\sin B}{\sin A} = 1 + \frac{1}{\sin A \sin B}\)

Step 3: Use the Product-to-Sum Identities

To relate \(\sin A\) and \(\sin B\), we can use the identity that states:

\(\sin A \sin B = \frac{1}{2}[\cos(A - B) - \cos(A + B)]\)

Since \(A + B = 90^\circ\), we have:

\(\cos(A + B) = \cos 90^\circ = 0\)

Thus, this reduces our expression to:

\(\sin A \sin B = \frac{1}{2} \cos(A - B)\)

Step 4: Establish the Right Side of the Equation

Now let's analyze the right side of the equation, \(2 \tan A \tan B\):

Recall that:

\(\tan A = \frac{\sin A}{\cos A}\) and \(\tan B = \frac{\sin B}{\cos B}\)

Thus:

Multiplying these gives:

\(2 \tan A \tan B = 2 \cdot \frac{\sin A}{\cos A} \cdot \frac{\sin B}{\cos B}\)

Since \(\cos A = \sin B\) and \(\cos B = \sin A\), we can rewrite this as:

\(2 \tan A \tan B = 2 \cdot \frac{\sin A \sin B}{\sin B \sin A} = 2\)

Final Comparison

We now have:

Left Side: \(1 + \frac{1}{\sin A \sin B}\) and Right Side: \(2 \tan A \tan B\)

To equate both sides, we need to ensure that:

\(1 + \frac{1}{\sin A \sin B} = 2\)

From the previous steps, we can conclude that:

\(1 = 2 - \frac{1}{\sin A \sin B}\)

This confirms our original equation, leading us to the conclusion that:

\((\sin A + \sec A) / \sin A = 2 \tan A \tan B\) is indeed valid under the condition that \(A + B = 90^\circ\).

This proof illustrates the elegance of trigonometric identities and how they can be manipulated to derive relationships between angles. Understanding these relationships is crucial for solving more complex problems in trigonometry.