If angle C of a triangle ABC is a right angle where a, b and c are the sides opposite to the angles A, B and C respectively then what is tan A + tan B equal to?
Finding tan A + tan B in a Right Angle Triangle
The problem asks us to find the value of \( \tan A + \tan B \) in a triangle ABC where angle C is a right angle (90 degrees). The sides opposite to angles A, B, and C are given as a, b, and c respectively.
In a right-angled triangle, the trigonometric ratios like tangent are defined based on the lengths of the sides relative to an angle.
For a right-angled triangle ABC with angle C = 90 degrees:
- The side opposite to angle A is a.
- The side adjacent to angle A is b.
- The hypotenuse (opposite to angle C) is c.
The definition of the tangent of an angle in a right-angled triangle is:
\( \tan(\text{angle}) = \frac{\text{Opposite Side}}{\text{Adjacent Side}} \)
Calculating tan A
For angle A, the opposite side is 'a' and the adjacent side is 'b'.
So, \( \tan A = \frac{a}{b} \)
Calculating tan B
For angle B, the opposite side is 'b' and the adjacent side is 'a'.
So, \( \tan B = \frac{b}{a} \)
Finding tan A + tan B
Now we need to find the sum of \( \tan A \) and \( \tan B \):
\( \tan A + \tan B = \frac{a}{b} + \frac{b}{a} \)
To add these fractions, we find a common denominator, which is ab:
\( \tan A + \tan B = \frac{a \times a}{b \times a} + \frac{b \times b}{a \times b} \)
\( \tan A + \tan B = \frac{a^2}{ab} + \frac{b^2}{ab} \)
Combine the numerators over the common denominator:
\( \tan A + \tan B = \frac{a^2 + b^2}{ab} \)
Using the Pythagorean Theorem
Since triangle ABC is a right-angled triangle with the right angle at C, the Pythagorean theorem applies. The theorem states that the square of the hypotenuse (c) is equal to the sum of the squares of the other two sides (a and b).
\( a^2 + b^2 = c^2 \)
We can substitute \( a^2 + b^2 \) with \( c^2 \) in our expression for \( \tan A + \tan B \):
\( \tan A + \tan B = \frac{c^2}{ab} \)
Therefore, \( \tan A + \tan B \) is equal to \( \frac{c^2}{ab} \).
Conclusion
In a right-angled triangle ABC with angle C = 90 degrees and sides a, b, c opposite to angles A, B, C respectively, the sum of \( \tan A + \tan B \) is found by first determining \( \tan A = a/b \) and \( \tan B = b/a \). Adding these gives \( \frac{a^2 + b^2}{ab} \). By the Pythagorean theorem, \( a^2 + b^2 = c^2 \), leading to the final result \( \frac{c^2}{ab} \).
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