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Consider the following statements: $\quad[2016-I]$
1) If $\mathrm{ABC}$ is an equilateral triangle, then $3 \tan (A+B) \tan C=1$
2) If $\mathrm{ABC}$ is a triangle in which $\mathrm{A}=78^{\circ}, \mathrm{B}=66^{\circ}$, then $\tan \left(\frac{A}{2}+C\right) < \tan A$
3) If $\mathrm{ABC}$ is any triangle, then
$$
\tan \left(\frac{\mathrm{A}+\mathrm{B}}{2}\right) \sin \left(\frac{\mathrm{C}}{2}\right) < \cos \left(\frac{\mathrm{C}}{2}\right)
$$
Which of the above statements is/are correct?
Options:
1) If $\mathrm{ABC}$ is an equilateral triangle, then $3 \tan (A+B) \tan C=1$
2) If $\mathrm{ABC}$ is a triangle in which $\mathrm{A}=78^{\circ}, \mathrm{B}=66^{\circ}$, then $\tan \left(\frac{A}{2}+C\right) < \tan A$
3) If $\mathrm{ABC}$ is any triangle, then
$$
\tan \left(\frac{\mathrm{A}+\mathrm{B}}{2}\right) \sin \left(\frac{\mathrm{C}}{2}\right) < \cos \left(\frac{\mathrm{C}}{2}\right)
$$
Which of the above statements is/are correct?
Solution:
2080 Upvotes
Verified Answer
The correct answer is:
2 only
$\because A B C$ is an equilateral triangle. $\therefore A=B=C=60^{\circ}$
L.H.S. $=3 \tan (A+B) \tan C$
$=3 \tan 120^{\circ} \tan 60^{\circ}$
$\quad=3(-\sqrt{3})(\sqrt{3})$
$\quad=-9 \neq 1$
Hence statement $(1)$ is incorrect. Statement-2 $A B C$ is a triangle such that $A=78^{\circ}$ and $B=66^{\circ}$ $C=180-(78+66)=180-144=36^{\circ}$
$\frac{A}{2}+C=\frac{78}{2}+36^{\circ}$
$=39+36=75^{\circ}$
$\tan \left(\frac{A}{2}+C\right) < \tan A$
$\Rightarrow \tan 75^{\circ} < \tan 78^{\circ}$
Hence statement $(2)$ is correct. Statement (3) In a triangle $\mathrm{ABC}$ $A+B+C=180^{\circ}$
$A+B=180^{\circ}-C$
$\frac{A+B}{2}=\frac{180^{\circ}-C}{2}$
$\Rightarrow \frac{A+B}{2}=90-\frac{C}{2}$
$\Rightarrow \tan \left(\frac{A+B}{2}\right)=\tan \left(90-\frac{C}{2}\right)$
$\Rightarrow \tan \left(\frac{A+B}{2}\right)=\cot \frac{C}{2}$
$\tan \left(\frac{A+B}{2}\right) \cdot \sin \frac{C}{2}=\cot \frac{C}{2} \cdot \sin \frac{C}{2}=\cos \frac{C}{2}$
$\Rightarrow \tan \left(\frac{A+B}{2}\right) \cdot \sin \frac{C}{2}=\cos \frac{C}{2}$
We can see that statement $(3)$ is not correct.
Hence only 2 nd statement is correct.
L.H.S. $=3 \tan (A+B) \tan C$
$=3 \tan 120^{\circ} \tan 60^{\circ}$
$\quad=3(-\sqrt{3})(\sqrt{3})$
$\quad=-9 \neq 1$
Hence statement $(1)$ is incorrect. Statement-2 $A B C$ is a triangle such that $A=78^{\circ}$ and $B=66^{\circ}$ $C=180-(78+66)=180-144=36^{\circ}$
$\frac{A}{2}+C=\frac{78}{2}+36^{\circ}$
$=39+36=75^{\circ}$
$\tan \left(\frac{A}{2}+C\right) < \tan A$
$\Rightarrow \tan 75^{\circ} < \tan 78^{\circ}$
Hence statement $(2)$ is correct. Statement (3) In a triangle $\mathrm{ABC}$ $A+B+C=180^{\circ}$
$A+B=180^{\circ}-C$
$\frac{A+B}{2}=\frac{180^{\circ}-C}{2}$
$\Rightarrow \frac{A+B}{2}=90-\frac{C}{2}$
$\Rightarrow \tan \left(\frac{A+B}{2}\right)=\tan \left(90-\frac{C}{2}\right)$
$\Rightarrow \tan \left(\frac{A+B}{2}\right)=\cot \frac{C}{2}$
$\tan \left(\frac{A+B}{2}\right) \cdot \sin \frac{C}{2}=\cot \frac{C}{2} \cdot \sin \frac{C}{2}=\cos \frac{C}{2}$
$\Rightarrow \tan \left(\frac{A+B}{2}\right) \cdot \sin \frac{C}{2}=\cos \frac{C}{2}$
We can see that statement $(3)$ is not correct.
Hence only 2 nd statement is correct.
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