In a triangle A B C, with usual notations, ≤ft(A+B/2) · ≤ft(A-B/2)=

Correct Answer is : a−ba+ba-b/a+ba+ba−b

Correct Answer is : a−ba+ba-b/a+ba+ba−b

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MHTCET 26 Apr 2025 Shift 1 Paper

Question : 120 of 150

Marks: +1, -0

In a triangle

A B C

\cot \left(\frac{A+B}{2}\right) \cdot \tan \left(\frac{A-B}{2}\right)=

Solution:

Concept:

Use Napier's analogy for tangents and the relation

A+B=\pi-C

Explanation:

In any triangle,

A+B+C=\pi

, so

\frac{A+B}{2}=\frac{\pi-C}{2}

Therefore,

\cot\left(\frac{A+B}{2}\right)=\cot\left(\frac{\pi}{2}-\frac{C}{2}\right)=\tan\left(\frac{C}{2}\right)

Using Napier's analogy,

\tan\left(\frac{A-B}{2}\right)=\frac{a-b}{a+b}\,\cot\left(\frac{C}{2}\right)

Multiplying both results:

\cot\left(\frac{A+B}{2}\right)\tan\left(\frac{A-B}{2}\right)=\tan\left(\frac{C}{2}\right)\cdot \frac{a-b}{a+b}\,\cot\left(\frac{C}{2}\right)=\frac{a-b}{a+b}

Answer:

\frac{a-b}{a+b}

which is option B.

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