Understanding the Problem: Trigonometry in a Right Triangle

The problem asks us to find the value of a trigonometric expression involving angles A and C in a right-angled triangle ABC, where the right angle is at B. We are given the value of tan A.

In a right-angled triangle ABC, with the right angle at B, the sum of angles A and C is 90 degrees (\(\text{A} + \text{C} = 90^\circ\)). This means angles A and C are complementary.

We are given that \( \text{tan A} = \frac{1}{2} \).

Finding Side Ratios from tan A

The tangent of an angle in a right-angled triangle is defined as the ratio of the side opposite the angle to the side adjacent to the angle.

For angle A:

• Opposite side is BC
• Adjacent side is AB

So, \( \text{tan A} = \frac{\text{BC}}{\text{AB}} \). Given \( \text{tan A} = \frac{1}{2} \), we can say that \( \frac{\text{BC}}{\text{AB}} = \frac{1}{2} \). We can assume the lengths of the sides BC and AB are in the ratio 1:2. Let BC = k and AB = 2k for some positive constant k.

Calculating the Hypotenuse AC

Using the Pythagorean theorem in the right-angled triangle ABC:

\( \text{AC}^2 = \text{AB}^2 + \text{BC}^2 \)

Substitute the values AB = 2k and BC = k:

\( \text{AC}^2 = (2k)^2 + k^2 \)

\( \text{AC}^2 = 4k^2 + k^2 \)

\( \text{AC}^2 = 5k^2 \)

\( \text{AC} = \sqrt{5k^2} = k\sqrt{5} \)

So, the lengths of the sides are AB = 2k, BC = k, and AC = \(k\sqrt{5}\).

Determining Sine and Cosine Values for Angles A and C

Now we can find the sine and cosine values for angles A and C using the side lengths.

For angle A:

• \( \text{sin A} = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{\text{BC}}{\text{AC}} = \frac{k}{k\sqrt{5}} = \frac{1}{\sqrt{5}} \)
• \( \text{cos A} = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{\text{AB}}{\text{AC}} = \frac{2k}{k\sqrt{5}} = \frac{2}{\sqrt{5}} \)

For angle C:

• \( \text{sin C} = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{\text{AB}}{\text{AC}} = \frac{2k}{k\sqrt{5}} = \frac{2}{\sqrt{5}} \)
• \( \text{cos C} = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{\text{BC}}{\text{AC}} = \frac{k}{k\sqrt{5}} = \frac{1}{\sqrt{5}} \)

Notice that since A and C are complementary, \( \text{sin C} = \text{cos A} \) and \( \text{cos C} = \text{sin A} \), which matches our calculations.

Evaluating the Trigonometric Expression

The expression we need to evaluate is \( \frac{\text{sin A}(\text{cos C} + \text{cos A})}{\text{cos C} (\text{sin C} - \text{sin A})} \).

Substitute the values we found:

• \( \text{sin A} = \frac{1}{\sqrt{5}} \)
• \( \text{cos A} = \frac{2}{\sqrt{5}} \)
• \( \text{sin C} = \frac{2}{\sqrt{5}} \)
• \( \text{cos C} = \frac{1}{\sqrt{5}} \)

Expression \( = \frac{\frac{1}{\sqrt{5}} \left( \frac{1}{\sqrt{5}} + \frac{2}{\sqrt{5}} \right)}{\frac{1}{\sqrt{5}} \left( \frac{2}{\sqrt{5}} - \frac{1}{\sqrt{5}} \right)} \)

Simplify the terms inside the parentheses:

• \( \text{cos C} + \text{cos A} = \frac{1}{\sqrt{5}} + \frac{2}{\sqrt{5}} = \frac{1+2}{\sqrt{5}} = \frac{3}{\sqrt{5}} \)
• \( \text{sin C} - \text{sin A} = \frac{2}{\sqrt{5}} - \frac{1}{\sqrt{5}} = \frac{2-1}{\sqrt{5}} = \frac{1}{\sqrt{5}} \)

Now substitute these back into the expression:

Expression \( = \frac{\frac{1}{\sqrt{5}} \left( \frac{3}{\sqrt{5}} \right)}{\frac{1}{\sqrt{5}} \left( \frac{1}{\sqrt{5}} \right)} \)

Multiply the terms in the numerator and the denominator:

Numerator \( = \frac{1}{\sqrt{5}} \times \frac{3}{\sqrt{5}} = \frac{1 \times 3}{\sqrt{5} \times \sqrt{5}} = \frac{3}{5} \)

Denominator \( = \frac{1}{\sqrt{5}} \times \frac{1}{\sqrt{5}} = \frac{1 \times 1}{\sqrt{5} \times \sqrt{5}} = \frac{1}{5} \)

Now divide the numerator by the denominator:

Expression \( = \frac{\frac{3}{5}}{\frac{1}{5}} = \frac{3}{5} \div \frac{1}{5} = \frac{3}{5} \times \frac{5}{1} \)

Cancel out the 5s:

Expression \( = 3 \)

Conclusion

The value of the given trigonometric expression is 3.

Revision Table: Right Triangle Trigonometry

Additional Information: Complementary Angles in Trigonometry

In a right-angled triangle, the two acute angles are always complementary. If these angles are A and C, then \( \text{A} + \text{C} = 90^\circ \). This relationship leads to important identities:

• \( \text{sin A} = \text{cos (90° - A)} = \text{cos C} \)
• \( \text{cos A} = \text{sin (90° - A)} = \text{sin C} \)
• \( \text{tan A} = \text{cot (90° - A)} = \text{cot C} \)
• \( \text{cot A} = \text{tan (90° - A)} = \text{tan C} \)
• \( \text{sec A} = \text{csc (90° - A)} = \text{csc C} \)
• \( \text{csc A} = \text{sec (90° - A)} = \text{sec C} \)

These identities can sometimes simplify trigonometric expressions involving complementary angles, as seen in this problem where \( \text{sin A} = \text{cos C} \) and \( \text{cos A} = \text{sin C} \).