Question

In$\Delta ABC,{\text{ }}2ac\sin \left( {\dfrac{{A - B + C}}{2}} \right)$is equal to:
$
  {\text{a}}{\text{. }}{a^2} + {b^2} - {c^2} \\
  {\text{b}}{\text{. }}{{\text{c}}^2} + {a^2} - {b^2} \\
  {\text{c}}{\text{. }}{{\text{b}}^2} - {a^2} - {c^2} \\
  {\text{d}}{\text{. }}{{\text{c}}^2} - {a^2} - {b^2} \\
$

Answer 1

Hint: Sum of all the angles in a triangle is $180^0$

We have to find out the value of
$2ac\sin \left( {\dfrac{{A - B + C}}{2}} \right)......................\left( 1 \right)$
As we know in triangle the sum of all angles is always equal to$180^\circ $
$
   \Rightarrow A + B + C = 180^\circ \\
   \Rightarrow A + C = 180^\circ - B \\
$
Therefore from equation 1
$2ac\sin \left( {\dfrac{{180^\circ - B - B}}{2}} \right) = 2ac\sin \left( {\dfrac{{180^\circ - 2B}}{2}} \right) = 2ac\sin \left( {90^\circ - B} \right)$
Now, we know that$\sin \left( {90^\circ - B} \right) = \cos B$, so, use this property
$ \Rightarrow 2ac\sin \left( {90^\circ - B} \right) = 2ac\cos B.................\left( 2 \right)$
Now as we know in any triangle the cosine of angle$B$is
i.e.$ \Rightarrow \cos B = \left( {\dfrac{{{a^2} + {c^2} - {b^2}}}{{2ac}}} \right)$
Therefore from equation 2
\[
   \Rightarrow 2ac\cos B = 2ac\left( {\dfrac{{{a^2} + {c^2} - {b^2}}}{{2ac}}} \right) \\
   = {a^2} + {c^2} - {b^2} \\
\]
Hence, option b is correct.

Note: -In such types of questions the key concept we have to remember is that the sum of all angles in any triangle is always$180^\circ $ and also remember the formula of cosine of any angle in a triangle, then simplify we will get the required answer.