Q62 / 160 · TS EAMCET 2020 (10 Sep Shift 1) · Mathematics
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 $\triangle P Q R$,
$\tan \frac{P}{2} \tan \frac{Q}{2}+\tan \frac{Q}{2} \tan \frac{R}{2}+\tan \frac{P}{2} \tan \frac{R}{2}=1$
The correct option among the following is
- Correct answerA(A) is true, (R) is true and (R) is the correct explanation for (A)
- B(A) is true, (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
Solution
Reason In $\triangle P Q R$,
$\begin{aligned} & P+Q+R=180^{\circ} \\ & \Rightarrow \quad \frac{P}{2}+\frac{Q}{2}+\frac{R}{2}=90^{\circ} \\ & \Rightarrow \quad \frac{P}{2}+\frac{Q}{2}=90^{\circ}-\frac{R}{2} \\ & \Rightarrow \quad \tan \left(\frac{P}{2}+\frac{Q}{2}\right)=\tan \left(90-\frac{R}{2}\right) \\ & \end{aligned}$
$\begin{aligned}
& \Rightarrow \quad \frac{\tan \frac{P}{2}+\tan \frac{Q}{2}}{1-\tan \frac{P}{2} \tan \frac{Q}{2}}=\cot \frac{R}{2} \\
& \Rightarrow \quad\left(\tan \frac{P}{2}+\tan \frac{Q}{2}\right) \tan \frac{R}{2}=1-\tan \frac{P}{2} \tan \frac{Q}{2} \\
& \Rightarrow \quad \tan \frac{P}{2} \tan \frac{Q}{2}+\tan \frac{Q}{2} \tan \frac{R}{2}+\tan \frac{R}{2} \tan \frac{P}{2}=1
\end{aligned}$
So, reason is true.
Assertion
We have,
$\begin{array}{lcc}
& A=15^{\circ}, B=17^{\circ}, C=13^{\circ} \\
\Rightarrow & A+B+C=45^{\circ} \\
\Rightarrow & 2 A+2 B+2 C=90^{\circ} \\
\therefore & \frac{P}{2}=2 A, \frac{Q}{2}=2 B, \frac{R}{2}=2 C \\
\therefore & \therefore \tan 2 A \tan 2 B+\tan 2 B \tan 2 C+\tan 2 C \tan 2 A=1
\end{array}$
$\begin{aligned} & \Rightarrow \frac{1}{\cot 2 A \cot 2 B}+\frac{1}{\cot 2 B \cot 2 C}+\frac{1}{\cot 2 C \cot 2 A}=1 \\ & \Rightarrow \frac{\cot 2 C+\cot 2 A+\cot 2 B}{\cot 2 A \cot 2 B \cot 2 C}=1 \\ & \Rightarrow \cot 2 A+\cot 2 B+\cot 2 C=\cot 2 A \cot 2 B \cot 2 C \\ & \therefore \text { Assertion is true. }\end{aligned}$
Hence, the correct answer is option: 'A'.