Question

Value of $\sin x\cos y+\sin z\sin y$ is equal to

• A. $\frac{1}{\sqrt{2}}$
• B. $\frac{\sqrt{3}}{2}$
• C. $0$
• D. $-\frac{1}{\sqrt{2}}$

Answer 1

Answer: (B)

$\frac{\sqrt{3}}{2}$

$\sum x=\frac{3\pi }{4},\sum \tan x=5,\tan x\tan y\tan z=1$

Using the expansion of $\tan (x+y+z),$

$\tan (x+y+z)=\frac{\sum \tan x-\tan x\tan y\tan z}{1-\tan x\tan y-\tan x\tan z-\tan y\tan z}$

Substituting the given values we get,

$\tan x\tan y+\tan x\tan z+\tan y\tan z=5$

Let $\tan x=a,\tan y=b,\tan z=c$

It can be observed that $a,b,c$ are roots of the equation,

$m^3-5m^3+5m^2-1=0$

Roots of which are,

$1,2\pm \sqrt{3}$

As $x<y<z$ and $\tan$ is increasing function

$\tan x=2-\sqrt{3},\tan y=1,\tan z=2+\sqrt{3}$

$\sin x=\frac{\sqrt{6}-\sqrt{2}}{4},\cos y=\frac{1}{\sqrt{2}}$

$\sin z=\frac{\sqrt{6}+\sqrt{2}}{4},\sin y=\frac{1}{\sqrt{2}}$

$\sin x\cos y+\sin z\sin y=\frac{\sqrt{3}}{2}$