Question:
\(BC=a,\; AC=b,\; AB=c\) are sides of \(\triangle ABC\) and \(\angle C\neq \frac{\pi}{2}\). Which of the following is not correct?
\(BC=a,\; AC=b,\; AB=c\) are sides of \(\triangle ABC\) and \(\angle C\neq \frac{\pi}{2}\). Which of the following is not correct?
Show Hint
Remember the important triangle identity:
\[
\frac{a-b}{a+b}
=
\tan\left(\frac{C}{2}\right)
\tan\left(\frac{A-B}{2}\right).
\]
It is frequently asked in trigonometry-based geometry problems.
Updated On: Jun 11, 2026
- \(\dfrac{a-b}{a+b}=\cot\left(\dfrac{A+B}{2}\right)\tan\left(\dfrac{A-B}{2}\right)\)
- \(\dfrac{a-b}{a+b}=\dfrac{\tan\left(\dfrac{A-B}{2}\right)}{\tan\left(\dfrac{C}{2}\right)}\)
- \(\dfrac{a-b}{a+b}=\dfrac{\sin A-\sin B}{\sin A+\sin B}\)
- \(\dfrac{a}{\sin A}=\dfrac{b}{\sin B}=\dfrac{c}{\sin C}\)
Show Solution
Verified By Collegedunia
The Correct Option is B
Solution and Explanation
Concept:
Using the Law of Sines,
\[
\frac{a}{\sin A}
=
\frac{b}{\sin B}
=
\frac{c}{\sin C}.
\]
Also,
\[
\frac{\sin A-\sin B}
{\sin A+\sin B}
=
\frac
{2\cos\left(\frac{A+B}{2}\right)
\sin\left(\frac{A-B}{2}\right)}
{2\sin\left(\frac{A+B}{2}\right)
\cos\left(\frac{A-B}{2}\right)}
\]
\[
=
\cot\left(\frac{A+B}{2}\right)
\tan\left(\frac{A-B}{2}\right).
\]
Since
\[
A+B=\pi-C,
\]
\[
\cot\left(\frac{A+B}{2}\right)
=
\tan\left(\frac{C}{2}\right).
\]
Therefore,
\[
\frac{a-b}{a+b}
=
\tan\left(\frac{C}{2}\right)
\tan\left(\frac{A-B}{2}\right).
\]
Hence option (B) is not equivalent to the standard identity because it contains division by \(\tan(C/2)\).
Thus the incorrect relation is
\[
\boxed{(B)}
\]
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