Question

Let $t_1=\cos A+\cos B-\cos (A+B)$ and $t_2=4\sin \frac{A}{2}.\sin \frac{B}{2}[2{\cos }^2\left(\frac{A+B}{4}\right)-1]$ then $t_1-t_2$ equals

• A. $0$
• B. $-1$
• C. $-2$
• D. $1$

Answer 1

The correct option is D $1$

As $t_1=\cos A+\cos B-\cos (A+B)$

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

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

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

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

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

And $t_2=4\sin \frac{A}{2}\sin \frac{B}{2}(2{\cos }^2\left(\frac{A+B}{2}\right)-1)$

Hence $t_1-t_2=1$