Question

I: $\cos x+\cos y=\frac{1}{3},\sin x+\sin y=\frac{1}{4}$, then $\cos (x+y)=\frac{-7}{25}$.
II. $\sin x+\sin y=\frac{1}{4},\sin x-\sin y=\frac{1}{5}$, then $4\cot \left(\frac{x-y}{2}\right)=5\cot \left(\frac{x+y}{2}\right)$

• 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

The correct option is B only II is true

I: $\cos x+\cos y=\frac{1}{3}........1$

$\sin x+\sin y=\frac{1}{4}........\left(2\right)$

Squaring and adding $\left(1\right)$ and $\left(2\right)$, we get

${\cos }^{2}x+{\cos }^{2}y+2\cos x\cos y+{\sin }^{2}x+{\sin }^{2}y+2\sin x\sin y=\frac{1}{9}+\frac{1}{16}......\left(3\right)$

$2+2\cos (x-y)=\frac{25}{144}$

$2+4{\cos }^2\frac{x-y}{2}-2=\frac{25}{144}$

${\cos }^2\frac{x-y}{2}=\frac{25}{144\times 4}$

Squaring and subtracting $\left(2\right)$ from $\left(1\right)$, we get

${\cos }^{2}x+{\cos }^{2}y+2\cos x\cos y-{\sin }^{2}x-{\sin }^{2}y-2\sin x\sin y=\frac{1}{9}-\frac{1}{16}$

$\cos 2x+\cos 2y+2\cos (x+y)=\frac{7}{144}$

$2\cos (x+y)\cos (x-y)+2\cos (x+y)=\frac{7}{144}$

$2\cos (x+y)\left(\cos \right(x-y)+1)=\frac{7}{144}$

$2\cos (x+y)\left(2{\cos }^2\right(x-y)/2-1+1)=\frac{7}{144}$

Substituting value of $\left(3\right)$

$2\cos (x+y)\cdot 2\cdot \frac{25}{144\times 4}=\frac{7}{144}$

$\cos (x+y)=\frac{7}{25}$

II: $\sin x+\sin y=\frac{1}{4}.........\left(1\right)$

$\sin x-\sin y=\frac{1}{5}..........\left(2\right)$

Dividing $\left(2\right)$ by $\left(1\right)$, we get

$\frac{\sin x+\sin y}{\sin x-\sin y}=\frac{5}{4}$

$\Rightarrow \frac{2\sin \left(\frac{x+y}{2}\right)\cos \left(\frac{x-y}{2}\right)}{2\cos \left(\frac{x+y}{2}\right)\sin \left(\frac{x-y}{2}\right)}=\frac{5}{4}$

$\Rightarrow \frac{\cot \left(\frac{x-y}{2}\right)}{\cot \left(\frac{x+y}{2}\right)}=\frac{5}{4}$

$\Rightarrow 4\cot \left(\frac{x-y}{2}\right)=5\cot \left(\frac{x+y}{2}\right)$