Question

lf $A+B+C={180}^{\circ }$. Match the following
List -I List -II

1. $\cos A+\cos B+\cos C$ a.$2+2\cos A\cos B\cos C$
2. ${\sin }^{2}A+{\sin }^{2}B+{\sin }^{2}C$ b.$4\cos \frac{A}{2}\cos \frac{B}{2}\cos \frac{C}{2}$

3. $\sin 2A+\sin 2B+\sin 2C$ c.$1+4\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2}$

4.$\sin A+\sin B+\sin C$ d. $4\sin A\sin B\sin C$

• A. $1-c,2-a,3-d,4-b$
• B. $1-d,2-b,3-a,4-c$
• C. $1-a,2-c,3-b,4-d$
• D. $1-b,2-d,3-c,4-a$

Answer 1

The correct option is A $1-c,2-a,3-d,4-b$

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

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

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

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

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

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

$=1+4\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2}$

B) ${\sin }^{2}A+{\sin }^{2}B+{\sin }^{2}C$

$=\frac{1-\cos 2A}{2}+\frac{1-\cos 2B}{2}+\frac{1-\cos 2B}{2}$

$=\frac{3}{2}-\frac{1}{2}(\cos 2A+\cos 2B+\cos 2C)$

$=\frac{3}{2}-\frac{1}{2}\left(2\cos \right(A+B\left)\cos \right(A-B)+\cos (2\pi -2(A+B)\left)\right)$

$=\frac{3}{2}-\frac{1}{2}\left(2\cos \right(A-B\left)\cos \right(A-B)+{\cos }^2(A+B)-1)$

$=2+2\cos A\cos B\cos C$

C) $\sin 2A+\sin 2B+\sin 2C$

$=2\sin (A+B)\cos (A-B)+\sin (2\pi -2(A+B\left)\right)$

$=2\sin (A+B)\cos (A-B)-{\sin }^2(A+B)$

$=2\sin (A+B)\cos (A-B)-2\sin (A+B)\cos (A-B)$

$=2\sin (A+B)\left(\cos \right(A-B\left)\right)-\cos (A+B)$

$=4\sin A\sin B\sin C$

D) $\sin A+\sin B+\sin C$

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

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

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

$=4\cos \frac{A}{2}\cos \frac{B}{2}\cos \frac{C}{2}$