Question

In $\Delta$ $ABC$ the sides opposite to angles $A,B,C$ are denoted by $a,b,c$ respectively.

The expression $\frac{(a+b+c)(b+c-a)(c+a-b)(a+b-c)}{4b^2{c}^2}$ is equal to

• A. $co{s}^{2}A$
• B. $si{n}^{2}A$
• C. cos A cos B cos C
• D. none of these

Answer 1

The correct option is B $si{n}^{2}A$

Simplifying the given expression, we get
$\frac{\left(\right(b+c{)}^2-a^2\left)\right(a-(b-c)\left)\right(a+(b-c))}{4b^2{c}^2}$
$=\frac{\left(\right(b+c{)}^2-a^2\left)\right(a^2-(b-c{)}^2)}{4b^2.c^2}$
$=\frac{b^2+c^2-a^2+2bc}{2bc}.-\left(\frac{b^2+c^2-a^2-2bc}{2bc}\right)$
$=(cosA+1)[-(cosA-1\left)\right]$ $[\because cosA=\frac{b^2+c^2-a^2}{2bc}]$
$=(1-cosA)(1+cosA)$
$=1-co{s}^{2}A$
$=si{n}^{2}A$.