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-angled triangle \( \triangle ABC \), right-angled at \( C \), we have:

\[ \tan A = \frac{a}{c}, \quad \tan B = \frac{b}{a} \]

Therefore,

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

Using the trigonometric identities in a right-angled triangle:

\[ a = c \sin A, \quad b = c \cos A, \quad \tan A = \frac{a}{b}, \quad \tan B = \frac{b}{a} \] \[ \tan A + \tan B = \frac{a}{b} + \frac{b}{a} = \frac{a^2 + b^2}{ab} \]

Since the triangle is right-angled at \( C \), we have \( a^2 + b^2 = c^2 \) (Pythagorean theorem).

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

So the correct expression for \( \tan A + \tan B \) is indeed \( \frac{c^2}{ab} \).

Final Answer: The final answer is \( {\frac{c^2}{ab}} \).