Question

I: lf $A+B+C={180}^{\circ }$, then ${\sin }^{2}A+{\sin }^{2}B+{\sin }^{2}C=2(1+\cos A\cos B\cos C)$
II : lf $A+B+C={180}^{\circ }$, then ${\cos }^{2}A+{\cos }^{2}B+{\cos }^{2}C=1-2\cos A\cos B\cos C$

• A. only I is true
• B. only II is true
• C. Both I and II are true
• D. Neither I nor II are true

Answer 1

Answer: (C)

Both I and II are true

$P={\sin }^{2}A+{\sin }^{2}B+{\sin }^{2}C$

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

Using $\cos 2x=2{\cos }^{2}x-1\Rightarrow {\cos }^{2}x=\frac{\cos 2x+1}{2}$

We get

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

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

Now from $A+B+C=180\Rightarrow 2C=360-2(A+B)$

$P=3-\frac{1}{2}\left(2\cos \right(A+B\left)\cos \right(A-B)+\cos (360-2(A+B))-3)$

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

$=3-\frac{1}{2}\left(2\cos \right(A+B\left)\right(\cos (A-B)+\cos (A+B))-4)$

$=3-\frac{1}{2}\left(2\cos \right(180-C\left)\right(2\cos A\cos B)-4)$

$=2\cos A\cos B\cos C+1$

II: Using $\cos 2x=2{\cos }^{2}x-1\Rightarrow {\cos }^{2}x=\frac{cos2x+1}{2}$

${\cos }^{2}A+{\cos }^{2}B+{\cos }^{2}C$

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

$A+B+C=180\Rightarrow 2C=360-2(A+B)$

$=\frac{1}{2}\left(2\cos \right(A+B\left)\cos \right(A-B)+\cos (360-2(A+B))+3)$

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

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

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

$=\frac{1}{2}\left(2\cos \right(180-C\left)\right(2\cos A\cos B)+2)$

$=-2\cos A\cos B\cos C+1$