Question

If $2\tan {10}^o+\tan {50}^o=2x,\tan {20}^0+\tan {50}^o=2y,2\tan {10}^o+\tan {70}^o=2w$ and $\tan {20}^o+\tan {70}^o=2z$, then which of the following is/are true ?

• A. $z>w>y>x$
• B. $w=x+y$
• C. $2y=z$
• D. $z+x=w+y$

Answer 1

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

The correct options are
A $z>w>y>x$
B $w=x+y$
C $2y=z$
D $z+x=w+y$

Consider $\tan {20}^0-2\tan {10}^0=\frac{\sin {20}^0\cos {10}^0-2\sin {10}^0\cos {20}^0}{\cos {10}^0\cos {20}^0}$$=\frac{(\sin {30}^0+\sin {10}^0)-2\left(\sin {30}^0-\sin {10}^0\right)}{2\cos {10}^0\cos {20}^0}$
$=\frac{3\sin {10}^0-\sin {30}^0}{2\cos {10}^0\cos {20}^0}=\frac{4{\sin }^3{10}^0}{2\cos {10}^0\cos {20}^0}>0$
$\Rightarrow \tan {20}^0>2\tan {10}^0$
$\therefore y>x$ and $z>w$ ...(i)
Now $2w-2y$
$=2\tan {10}^0+\tan {70}^0-\tan {20}^0-\tan {50}^0$
$=2\tan {10}^0+\tan {20}^0\tan {50}^0\tan {70}^0$
$=2\tan {10}^0+\tan {50}^0=2x$ which is positive.
$\therefore w=x+y$ and $w>y$
Combining (i) and (ii) $z>w>y>x$
$2y-z=\tan {20}^0+\tan {50}^0-\frac{1}{2}\tan {20}^0-\frac{1}{2}\tan {70}^0$
$=\frac{1}{2}\tan {20}^0+\tan {50}^0-\frac{1}{2}\tan {70}^0$
$=\frac{1}{2}(\tan {20}^0+2\tan {50}^0-\tan {70}^0)$
$=\frac{1}{2}(\frac{\sin {70}^0}{\cos {20}^0\cos {50}^0}-\frac{\sin {20}^0}{\cos {50}^0\cos {70}^0})$
$=\frac{\sin {140}^0-\sin {40}^0}{2\cos {20}^0\cos {50}^0\cos {70}^0}=0$