Question

If areas of ${△}^{\prime }$'s $AMN,BNL$ and $CLM$ are ${△}_1,{△}_2$ and ${△}_3$ respectively, then the value of ${△}_1+{△}_2+{△}_3$ is

• A. $△(2+2\cos A\cos B\cos C)$
• B. $△(2+2\sin A\sin B\sin C)$
• C. $△(1-2\cos A\cos B\cos C)$
• D. $△(1-2\sin A\sin B\sin C)$

Answer 1

The correct option is C $△(1-2\cos A\cos B\cos C)$

In $△ABM,$
$\cos A=\frac{AM}{AB}=\frac{AM}{c}$
$\therefore AM=c\cos A$
Similarly, $AN=b\cos A$
${△}_1=$ Area of $△AMN$
$=\frac{1}{2}.AM.AN.\sin A$
$=\frac{1}{2}\left(c\cos A\right)\left(b\cos A\right).\left(\sin A\right)$
$=\left(\frac{1}{2}bc\sin A\right){\cos }^{2}A$
$=△{\cos }^{2}A$ where $△=\frac{1}{2}bc\sin A$
${△}_2=$Area of $△BNL$
$=\frac{1}{2}.BL.BN.\sin A$
$=\frac{1}{2}\left(a\cos B\right)\left(c\cos B\right)\sin B$
$=\left(\frac{1}{2}ac\sin B\right){\cos }^{2}B$
$=△{\cos }^{2}B$ where $△=\frac{1}{2}ac\sin B$
${△}_3=$Area of $△CMN$
$=\frac{1}{2}.CM.CL$
$=\frac{1}{2}\left(b\cos C\right)\left(a\cos C\right)\sin C$
$=\left(\frac{1}{2}ab\sin B\right){\cos }^{2}C$
$=△{\cos }^{2}B$ where $△=\frac{1}{2}ab\sin B$
$\therefore {△}_1+{△}_2+{△}_3=△({\cos }^{2}A+{\cos }^{2}B+{\cos }^{2}C)$
Consider, ${\cos }^{2}A+{\cos }^{2}B+{\cos }^{2}C$
We know that $\cos 2A=2{\cos }^{2}A-1$
or ${\cos }^{2}A=\frac{1+\cos 2A}{2}$
$\Rightarrow {\cos }^{2}A+{\cos }^{2}B+{\cos }^{2}C$
$=\frac{1+\cos 2A}{2}+\frac{1+\cos 2B}{2}+\frac{1+\cos 2C}{2}$
$=\frac{3}{2}+\frac{1}{2}[\cos 2A+\cos 2B+\cos 2C]$
$=\frac{3}{2}+\frac{1}{2}[2\cos (A+B)\cos (A-B)+2{\cos }^{2}C-1]$
$=\frac{3}{2}+\cos (\pi -C)\cos (A-B)+{\cos }^{2}C-\frac{1}{2}$
$=1-\cos C\cos (A-B)+{\cos }^{2}C$
$=1-\cos C[\cos (A-B)+\cos C]$
$=1-\cos C[\cos (A-B)+\cos (\pi -(A+B))]$
$=1-\cos C[\cos (A+B)+\cos (A+B)]$
Using transformation angle formula, $\sin C+\sin D=2\sin \left(\frac{C+D}{2}\right)\cos \left(\frac{C-D}{2}\right)$
$=1-\cos C\left[2\cos A\cos B\right]$
$=1-2\cos A\cos B\cos C$
substituting for ${\cos }^{2}A+{\cos }^{2}B+{\cos }^{2}C=1-2\cos A\cos B\cos C$ we get
$\therefore {△}_1+{△}_2+{△}_3=△(1-2\cos A\cos B\cos C)$