Question

If $\triangle ABC$ is right-angled at $C$, then the value of $\tan A + \tan B$ is:

• A. $a + b$
• B. $\frac{a^2}{bc}$
• C. $\frac{c^2}{ab}$
• D. $\frac{b^2}{ac}$

Answer 1

The Correct Option is C

Given: Right-angled triangle \( \Delta ABC \) with right angle at \( C \).

Step 1: Label the sides.

Let:

• \( AB = c \) (hypotenuse)
• \( BC = a \)
• \( AC = b \)

 

Step 2: Express \(\tan A\) and \(\tan B\).

\[ \tan A = \frac{\text{opposite}}{\text{adjacent}} = \frac{a}{b} \] \[ \tan B = \frac{\text{opposite}}{\text{adjacent}} = \frac{b}{a} \]

Step 3: Compute \(\tan A + \tan B\).

\[ \tan A + \tan B = \frac{a}{b} + \frac{b}{a} = \frac{a^2 + b^2}{ab} \]

Step 4: Apply Pythagorean theorem.

\[ a^2 + b^2 = c^2 \Rightarrow \tan A + \tan B = \frac{c^2}{ab} \]

Option Analysis:

(A) \( a + b \) - Incorrect (sum of sides, not tangent sum)

(B) \( \frac{a^2}{bc} \) - Incorrect form

(C) \( \frac{c^2}{ab} \) - Correct (matches our derivation)

(D) \( \frac{b^2}{ac} \) - Incorrect form

Final Answer: \(\boxed{C}\)

Answer 2

In a right triangle, $\tan A = \frac{BC}{AC}$ and $\tan B = \frac{AC}{BC}$.
Then: $\tan A + \tan B = \frac{BC}{AC} + \frac{AC}{BC} = \frac{BC^2 + AC^2}{BC \cdot AC}$. 
Using the Pythagorean theorem, $BC^2 + AC^2 = AB^2 = c^2$. 
Hence: $\tan A + \tan B = \frac{c^2}{ab}$.