Question

Assertion (A) : If $A=15^{\circ}, B=17^{\circ}$ and $C=13^{\circ}$ then $\cot 2 A+\cot 2 B+\cot 2 C=\cot 2 A \cot 2 B \cot 2 C$
Reason (R): In a $\triangle PQR$, $\text{Tan} \frac{P}{2} \text{Tan} \frac{Q}{2}+\text{Tan} \frac{Q}{2} \text{Tan} \frac{R}{2}+\text{Tan} \frac{P}{2} \text{Tan} \frac{R}{2}=1$

• A. (A) is true, (R) is true and (R) is the correct explanation for (A)
• B. (A) is tine, (R) is true but (R) is not the correct explanation for (A)
• C. (A) is true but (R) is false
• D. (A) is false but (R) is true

Answer 1

Answer: (A)

Reason In $\triangle P Q R$
$P+Q+R=180^{\circ}$
$\Rightarrow \frac{P}{2}+\frac{Q}{2}+\frac{R}{2}=90^{\circ}$
$\Rightarrow \frac{P}{2}+\frac{Q}{2}=90^{\circ}-\frac{R}{2} $
$\Rightarrow \tan \left(\frac{P}{2}+\frac{Q}{2}\right)=\tan \left(90-\frac{R}{2}\right) $
$\Rightarrow \frac{\tan \frac{P}{2}+\tan \frac{Q}{2}}{1-\tan \frac{P}{2} \tan \frac{Q}{2}}=\cot \frac{R}{2}$
$\Rightarrow \left(\tan \frac{P}{2}+\tan \frac{Q}{2}\right) \tan \frac{R}{2}=1-\tan \frac{P}{2} \tan \frac{Q}{2} $
$\Rightarrow \tan \frac{P}{2} \tan \frac{Q}{2}+\tan \frac{Q}{2} \tan \frac{R}{2}+\tan \frac{R}{2} \tan \frac{P}{2}=1$
So, reason is true.
Assertion
We have,
$A=15^{\circ}, B=17^{\circ}, C=13^{\circ}$
$\Rightarrow A+B+C=45^{\circ}$
$\Rightarrow 2 A+2 B+2 C=90^{\circ}$
$\therefore \frac{P}{2}=2 A, \frac{Q}{2}=2 B, \frac{R}{2}=2 C$
$\therefore \tan 2 A \tan 2 B+\tan 2 B \tan 2 C+\tan 2 C \tan 2 A=1$
$\Rightarrow \frac{1}{\cot 2 A \cot 2 B}+\frac{1}{\cot 2 B \cot 2 C}+\frac{1}{\cot 2 C \cot 2 A}=1$
$\Rightarrow \frac{\cot 2 C+\cot 2 A+\cot 2 B}{\cot 2 A \cot 2 B \cot 2 C}=1$
$\Rightarrow \cot 2 A+\cot 2 B+\cot 2 C$
$=\cot 2 A \cot 2 B \cot 2 C$
$\therefore $ Assertion is true.