Finding sin A when sec A + tan A is Given

This question asks us to find the value of \( \sin A \) given the relationship \( \sec A + \tan A = 5 \). We can solve this trigonometric problem by using fundamental trigonometric identities and solving a system of equations.

Using Fundamental Trigonometric Identities

A key trigonometric identity relates secant and tangent: \( \sec^2 A - \tan^2 A = 1 \). This identity is derived from the Pythagorean identity \( 1 + \tan^2 A = \sec^2 A \).

The identity \( \sec^2 A - \tan^2 A = 1 \) can be factored as a difference of squares:

\( (\sec A - \tan A)(\sec A + \tan A) = 1 \)

We are given that \( \sec A + \tan A = 5 \). Substituting this into the factored identity:

\( (\sec A - \tan A)(5) = 1 \)

Dividing both sides by 5, we get a second relationship between \( \sec A \) and \( \tan A \):

\( \sec A - \tan A = \frac{1}{5} \)

Solving for sec A and tan A

Now we have a system of two linear equations involving \( \sec A \) and \( \tan A \):

1. \( \sec A + \tan A = 5 \)
2. \( \sec A - \tan A = \frac{1}{5} \)

We can solve this system by adding or subtracting the equations.

Adding the two equations:

\( (\sec A + \tan A) + (\sec A - \tan A) = 5 + \frac{1}{5} \)

\( 2 \sec A = \frac{25}{5} + \frac{1}{5} \)

\( 2 \sec A = \frac{26}{5} \)

Divide by 2 to find \( \sec A \):

\( \sec A = \frac{26}{5} \times \frac{1}{2} = \frac{13}{5} \)

Subtracting the second equation from the first:

\( (\sec A + \tan A) - (\sec A - \tan A) = 5 - \frac{1}{5} \)

\( \sec A + \tan A - \sec A + \tan A = \frac{25}{5} - \frac{1}{5} \)

\( 2 \tan A = \frac{24}{5} \)

Divide by 2 to find \( \tan A \):

\( \tan A = \frac{24}{5} \times \frac{1}{2} = \frac{12}{5} \)

Finding sin A

We have found \( \sec A = \frac{13}{5} \) and \( \tan A = \frac{12}{5} \). We need to find \( \sin A \).

We know that \( \tan A = \frac{\sin A}{\cos A} \). Also, \( \sec A = \frac{1}{\cos A} \). From \( \sec A = \frac{13}{5} \), we can find \( \cos A \):

\( \cos A = \frac{1}{\sec A} = \frac{1}{13/5} = \frac{5}{13} \)

Now we can use the relationship \( \tan A = \frac{\sin A}{\cos A} \) to find \( \sin A \):

\( \sin A = \tan A \times \cos A \)

Substitute the values of \( \tan A \) and \( \cos A \):

\( \sin A = \frac{12}{5} \times \frac{5}{13} \)

\( \sin A = \frac{12 \times 5}{5 \times 13} \)

\( \sin A = \frac{12}{13} \)

Alternatively, since we have \( \tan A = \frac{12}{5} \), which is the ratio of the opposite side to the adjacent side in a right-angled triangle, we can construct a right triangle. If the opposite side is 12k and the adjacent side is 5k for some constant k (we can assume k=1 for simplicity in ratios), the hypotenuse would be calculated using the Pythagorean theorem:

\( \text{Hypotenuse} = \sqrt{(\text{Opposite})^2 + (\text{Adjacent})^2} = \sqrt{12^2 + 5^2} = \sqrt{144 + 25} = \sqrt{169} = 13 \)

So, in this right triangle, the sides are 12, 5, and 13. Now we can find \( \sin A \), which is the ratio of the opposite side to the hypotenuse:

\( \sin A = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{12}{13} \)

We can also check if \( \sec A + \tan A = 5 \) holds with these values:

\( \cos A = \frac{5}{13} \implies \sec A = \frac{13}{5} \)

\( \sec A + \tan A = \frac{13}{5} + \frac{12}{5} = \frac{13+12}{5} = \frac{25}{5} = 5 \)

This confirms our values for \( \sec A \), \( \tan A \), and \( \sin A \) are consistent with the given information.

Therefore, \( \sin A \) is equal to \( \frac{12}{13} \).

Revision Table: Trigonometric Ratios and Identities

Additional Information: Important Trigonometric Identities

Beyond \( \sec^2 A - \tan^2 A = 1 \), several other identities are crucial for solving trigonometry problems:

• Pythagorean Identities:
  • \( \sin^2 A + \cos^2 A = 1 \)
  • \( 1 + \tan^2 A = \sec^2 A \)
  • \( 1 + \cot^2 A = \csc^2 A \)
• Reciprocal Identities:
  • \( \sec A = \frac{1}{\cos A} \)
  • \( \csc A = \frac{1}{\sin A} \)
  • \( \cot A = \frac{1}{\tan A} \)
• Quotient Identities:
  • \( \tan A = \frac{\sin A}{\cos A} \)
  • \( \cot A = \frac{\cos A}{\sin A} \)

Understanding and applying these identities is key to simplifying expressions and solving equations in trigonometry. The identity \( \sec^2 A - \tan^2 A = 1 \) is particularly useful when dealing with sums or differences of secant and tangent, as shown in this problem.