Question

STATEMENT 1: In a $\Delta ABC,\cos A+\cos B+\cos C>1$, then the nature of the triangle cannot be determined.
STATEMENT 2: In a $\Delta ABC,\cos A+\cos B+\cos C=1+4\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2}$

• A. Statement-1 is true, Statement-2 is true,Statement-2 is the correct explanation of Statement- 1.
• B. Statement-1 is true, Statement-2 is true,Statement-2 is not the correct explanation for Statement-1.
• C. Statement-1 is true, Statement-2 is false.
• D. Statement-1 is false, Statement-2 is true.

Answer 1

Answer: (A)

Statement-1 is true, Statement-2 is true,Statement-2 is the correct explanation of Statement- 1.

$\cos A+\cos B+\cos C$

$=2\cos \left(\frac{A+B}{2}\right)\cos \left(\frac{A-B}{2}\right)-2{\sin }^2\frac{C}{2}+1$

$=2\sin \left(\frac{C}{2}\right)\cos \left(\frac{A-B}{2}\right)-2{\sin }^2\frac{C}{2}+1$

$=2\sin \left(\frac{C}{2}\right)[\cos \left(\frac{A-B}{2}\right)-\sin \left(\frac{C}{2}\right)]+1$

$=2\sin \left(\frac{C}{2}\right)[\cos \left(\frac{A-B}{2}\right)-\cos \left(\frac{A+B}{2}\right)]+1$

$=2\sin \left(\frac{C}{2}\right)\left[2\sin \frac{A}{2}\sin \frac{B}{2}\right]+1$

$=1+4\sin \left(\frac{C}{2}\right)\sin \left(\frac{A}{2}\right)\sin \left(\frac{B}{2}\right)$

Now

$\cos A+\cos B+\cos C>1$

$⟹1+4\sin \left(\frac{C}{2}\right)\sin \left(\frac{A}{2}\right)\sin \left(\frac{B}{2}\right)>1$

Or

$4\sin \left(\frac{C}{2}\right)\sin \left(\frac{A}{2}\right)\sin \left(\frac{B}{2}\right)>0$

Or

$\sin \left(\frac{C}{2}\right)\sin \left(\frac{A}{2}\right)\sin \left(\frac{B}{2}\right)>0$

Hence we cannot determine the nature of the triangle using this criterion only.