Question

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

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