Question

If $2a=2\tan {10}^{\circ }+\tan {50}^{\circ }$
$2b=\tan {20}^{\circ }+\tan {50}^{\circ }$
$2c=2\tan {10}^{\circ }+\tan {70}^{\circ }$
$2d=\tan {20}^{\circ }+\tan {70}^{\circ }$,
then which of the following is/are correct ?

• A. $a+d=b+c$
• B. $a+b=c$
• C. $a>b<c>d$
• D. $a<b<c<d$

Answer 1

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

The correct options are
A $a+d=b+c$
B $a+b=c$
D $a<b<c<d$

(A) $a+d=b+c\Rightarrow 2(a+d)=2(b+c)$
$\Rightarrow 2\tan {10}^{\circ }+\tan {50}^{\circ }+\tan {20}^{\circ }+\tan {70}^{\circ }=\tan {20}^{\circ }+\tan {50}^{\circ }+2\tan {10}^{\circ }+\tan {70}^{\circ }$

(B) $\tan {50}^{\circ }=\frac{1}{\tan {40}^{\circ }}=\frac{1-{\tan }^2{20}^{\circ }}{2\tan {20}^{\circ }}$
$\Rightarrow \tan {20}^{\circ }+2\tan {50}^{\circ }=\tan {70}^{\circ }$ $[\because \tan {20}^{\circ }=\cot {70}^{\circ }]$
$\Rightarrow 2a+2b=2c\Rightarrow a+b=c$

(D) $\tan {20}^{\circ }-2\tan {10}^{\circ }=\tan {20}^{\circ }{\tan }^2{10}^{\circ }>0$ $[\because \tan {20}^{\circ }=\frac{2\tan {10}^{\circ }}{1-{\tan }^2{10}^{\circ }}]$
$\Rightarrow \tan {20}^{\circ }>2\tan {10}^{\circ }$
$\Rightarrow b>a$ and $d>c$