Question

Let $0\le x,y,z\le \frac{\pi }{2}$ such that $\sin x\cdot \sin y\cdot \cos z=\frac{1}{4\sqrt{2}},\frac{{\sin }^{2}x\cdot {\sin }^{3}y}{\cos z}=\frac{1}{4\sqrt{2}}\text{and}\frac{\sin x\cdot {\sin }^{4}y}{{\cos }^{2}z}=\frac{1}{8}$, then which of the following is/are correct?

• A. $\sin x+\sin y=\frac{3}{2}$
• B. $\cos (x-y)=\frac{\sqrt{3}}{2}$
• C. $\tan (x+y+z)=\sqrt{3}-2$
• D. $\sec (y-z)=\sqrt{6}-\sqrt{2}$

Answer 1

Answer: (A), (C), (D)

The correct options are
A $\sin x+\sin y=\frac{3}{2}$
C $\tan (x+y+z)=\sqrt{3}-2$
D $\sec (y-z)=\sqrt{6}-\sqrt{2}$

Let $\sin x=a,\sin y=b,\&\cos z=c$

$\Rightarrow abc=\frac{1}{2\sqrt{2}},\frac{a^2{b}^3}{c}=\frac{1}{4\sqrt{2}}$ and $\frac{a{b}^4}{c^2}=\frac{1}{8}$

$3$ equations, $3$ unknowns
By solving, $a=1,b=\frac{1}{2},c=\frac{1}{\sqrt{2}}$
$\Rightarrow x=\frac{\pi }{2},y=\frac{\pi }{6},z=\frac{\pi }{4}$

$\Rightarrow \sin x+\sin y=\frac{3}{2}$
$\Rightarrow \cos (x-y)=\frac{1}{2}$
$\Rightarrow \tan (x+y+z)=\sqrt{3}-2$
$\Rightarrow \sec (y-z)=\sqrt{6}-\sqrt{2}$