Question

If A,B and C are the interior angles of a triangle ABC, then the value of $sin\left(\frac{B+C}{2}\right)\times cot\left(\frac{A+B-C}{2}\right)$ is equal to

• A.
  $Cos\frac{A}{2}$
• B.
  $tanC$
• C.
  $cos\frac{A}{2}\times tanC$
• D.
  $cosA\times tanC$

Answer 1

The correct option is C

$cos\frac{A}{2}\times tanC$
Given that A, B and C are the interior angles of $\Delta ABC.$ By angle sum property of triangles,$A+B+C={180}^{\circ }$

Consider: $sin\left(\frac{B+C}{2}\right)\times cot\left(\frac{A+B-C}{2}\right)=sin\left(\frac{{180}^{\circ }-A}{2}\right)\times cot\left(\frac{{180}^{\circ }-C-C}{2}\right)(\because A+B+C={180}^{\circ })=sin({90}^{\circ }-\frac{A}{2})\times cot({90}^{\circ }-C)=cos\frac{A}{2}\times tanC$

Hence, the correct answer is option c.