Question

In a \( \Delta ABC \), \( \angle B = 90^\circ \), then \( \tan^2\left(\frac{A}{2}\right) \) is

• A. $\frac{b-c}{b+c}$
• B. $\frac{b+c}{b-c}$
• C. $\frac{b-2c}{b+c}$
• D. None of these

Hint:

In a $Δ ABC$, $\angle B = 90circ$, then $\tan(A/2)$ is

Answer 1

The Correct Option is A

Step 1: Understanding the Concept
In a right-angled triangle at $B$, $\cos A = c/b$.
Step 2: Evaluation
Use the identity $\tan^2(A/2) = \frac{1-\cos A}{1+\cos A}$.
Step 3: Final Calculation
$\tan^2(A/2) = \frac{1 - c/b}{1 + c/b} = \frac{(b-c)/b}{(b+c)/b}$.
Step 4: Conclusion
The result is $\frac{b-c}{b+c}$.
Hence, the Answer is: (a)