Question

In $△ABC$, find the value of $\frac{1-co{s}^{2}A-co{s}^{2}B-co{s}^{2}C}{cosAcosBcosC}$

• A. $1$
• B. $2$
• C. $21$
• D. $35$

Answer 1

Answer: (B)

$2$

$A+B+C={180}^{\circ }$

$A+B={180}^{\circ }-C$

$cos(A+B)=cos({180}^{\circ }-C)=-cosC$

We know that

$cos2x=2co{s}^{2}x-1$

So,

$co{s}^{2}A=\frac{cos2A+1}{2}$

$co{s}^{2}B=\frac{cos2B+1}{2}$

$co{s}^{2}C=\frac{cos2C+1}{2}$

$cosx+cosy=2cos\left(\frac{x+y}{2}\right)cos\left(\frac{x-y}{2}\right)$

Therefore,

$1-co{s}^{2}A-co{s}^{2}B-co{s}^{2}C=1-\left(\frac{cos2A+1}{2}\right)-\left(\frac{cos2B+1}{2}\right)-\left(\frac{cos2C+1}{2}\right)$

$=\frac{-1-cos2A-cos2B-cos2C}{2}=-\frac{-1-\left(2cos\right(A+B\left)cos\right(A-B\left)\right)-cos2C}{2}$

$=\frac{-1-2(-cosC)cos(A-B)-cos2C}{2}=\frac{2cosCcos(A-B)-2co{s}^{2}C}{2}$

$=cosC\left(cos\right(A-B)-cosC)=2cosCsin\left(\frac{A-B+C}{2}\right)sin\left(\frac{C-A+B}{2}\right)$

$=2cosCsin({90}^{\circ }-B)sin({90}^{\circ }-A)=2cosAcosBcosC$

Hence,

L.H.S$=\frac{2cosAcosBcosC}{cosAcosBcosC}=2$