If cosAcosB-sinA/cosecB=4/5, (sinA/secB)-cosAsinB=5/13. Find 2/(cotA(1-tan^2A)). I need numeric value

We are given the following two trigonometric equations:

\[
\cos A \cos B - \frac{\sin A}{\csc B} = \frac{4}{5}
\]
\[
\frac{\sin A}{\sec B} - \cos A \sin B = \frac{5}{13}
\]

We are asked to find the numeric value of:

\[
\frac{2}{\cot A (1 - \tan^2 A)}
\]

### Step 1: Simplify the given expressions

#### Equation 1:
The first equation is:

\[
\cos A \cos B - \frac{\sin A}{\csc B} = \frac{4}{5}
\]

Since \(\csc B = \frac{1}{\sin B}\), the second term simplifies as:

\[
\cos A \cos B - \sin A \sin B = \frac{4}{5}
\]

This is a known trigonometric identity for \(\cos(A+B)\), so we get:

\[
\cos(A+B) = \frac{4}{5}
\]

#### Equation 2:
The second equation is:

\[
\frac{\sin A}{\sec B} - \cos A \sin B = \frac{5}{13}
\]

Since \(\sec B = \frac{1}{\cos B}\), this simplifies as:

\[
\sin A \cos B - \cos A \sin B = \frac{5}{13}
\]

This is a known trigonometric identity for \(\sin(A+B)\), so we get:

\[
\sin(A+B) = \frac{5}{13}
\]

### Step 2: Use Pythagoras' Theorem for Trigonometric Ratios
We now have:

\[
\cos(A+B) = \frac{4}{5}, \quad \sin(A+B) = \frac{5}{13}
\]

To find the tangent of \(A+B\), we use the identity:

\[
\tan(A+B) = \frac{\sin(A+B)}{\cos(A+B)}
\]

Substituting the values:

\[
\tan(A+B) = \frac{5/13}{4/5} = \frac{5}{13} \times \frac{5}{4} = \frac{25}{52}
\]

### Step 3: Find \(\frac{2}{\cot A (1 - \tan^2 A)}\)

We need to find:

\[
\frac{2}{\cot A (1 - \tan^2 A)}
\]

First, recall that \(\cot A = \frac{1}{\tan A}\) and \(\tan^2 A\) is the square of the tangent of angle \(A\).

Next, compute

If cosAcosB - sinA/cosecB = 4/5, (sinA/secB) - cosAsinB = 5/13. Find 2/(cotA(1-tan^2A)). I need numeric value.

Learn how to solve the trigonometric problem involving the identities cosAcosB - sinA/cosecB = 4/5 and sinA/secB - cosAsinB = 5/13. This step-by-step solution covers key concepts like cos(A + B), sin(A + B), and cotangent identities, ideal for students in Grades 10-12.

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