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Assertion (A) If $\alpha=12^{\circ}, \beta=15^{\circ}, \gamma=18^{\circ}$, then $\tan 2 \alpha \tan 2 \beta+\tan 2 \beta \tan 2 \gamma+\tan 2 \gamma \tan 2 \alpha=1$
Reason (R) In $\triangle A B C \cdot \tan \frac{A}{2} \tan \frac{B}{2}+\tan \frac{B}{2} \tan \frac{C}{2}$
$$
+\tan \frac{C}{2} \tan \frac{A}{2}=1
$$
Which of the following is true?
Options:
Reason (R) In $\triangle A B C \cdot \tan \frac{A}{2} \tan \frac{B}{2}+\tan \frac{B}{2} \tan \frac{C}{2}$
$$
+\tan \frac{C}{2} \tan \frac{A}{2}=1
$$
Which of the following is true?
Solution:
2815 Upvotes
Verified Answer
The correct answer is:
Both (A) and (R) are true and (R) is the correct explanation of (A)
$$
\begin{aligned}
& \text { } \alpha=12^{\circ}, \beta=15^{\circ}, \gamma=18^{\circ} \\
& \Rightarrow \tan 2 \alpha \tan 2 \beta+\tan 2 \beta \tan 2 \gamma \\
& +\tan 2 \gamma \tan 2 \alpha=1 \\
& \Rightarrow \tan 2 \alpha[\tan 2 \beta+\tan 2 \gamma]=1-\tan 2 \beta \tan 2 \gamma \\
& \Rightarrow \tan 2 \alpha\left[\frac{\tan 2 \beta+\tan 2 \gamma}{1+\tan 2 \beta \tan 2 \gamma}\right]=1 \\
& \Rightarrow \quad \tan 2 \alpha \cdot \tan (2 \beta+2 \gamma)=1 \\
& \Rightarrow \quad \tan (2 \beta+2 \gamma)=\cot 2 \alpha \\
& \Rightarrow \quad \tan \{30+36\}=\cot 24 \\
& \Rightarrow \quad \tan 66^{\circ}=\cot 24^{\circ} \\
& \Rightarrow \quad \tan 66^{\circ}=\tan \left(90^{\circ}-66^{\circ}\right) \\
& \Rightarrow \tan 66^{\circ}=\tan 66^{\circ} \text { which is true } \\
&
\end{aligned}
$$
$$
\begin{aligned}
\mathrm{R} \rightarrow & \tan \frac{A}{2} \tan \frac{B}{2}+\tan \frac{B}{2} \tan \frac{C}{2}=1-\tan \frac{C}{2} \tan \frac{A}{2} \\
& \frac{\tan \frac{B}{2}\left[\tan \frac{A}{2}+\tan \frac{C}{2}\right]}{1-\tan \frac{A}{2} \tan \frac{C}{2}}=1 \\
\Rightarrow & \tan \left[\frac{A}{2}+\frac{C}{2}\right]=\cot \frac{B}{2} \Rightarrow A+B+C=180 \\
\Rightarrow & \quad A+C=180-B \Rightarrow \frac{A+C}{2}=90-\frac{B}{2} \\
\Rightarrow & \tan \left[90^{\circ}-\frac{B}{2}\right]=\cot \frac{B}{2} \Rightarrow \cot \frac{B}{2}=\cot \frac{B}{2}
\end{aligned}
$$
$\mathrm{R}$ is true and it is correct explanation of $A$.
\begin{aligned}
& \text { } \alpha=12^{\circ}, \beta=15^{\circ}, \gamma=18^{\circ} \\
& \Rightarrow \tan 2 \alpha \tan 2 \beta+\tan 2 \beta \tan 2 \gamma \\
& +\tan 2 \gamma \tan 2 \alpha=1 \\
& \Rightarrow \tan 2 \alpha[\tan 2 \beta+\tan 2 \gamma]=1-\tan 2 \beta \tan 2 \gamma \\
& \Rightarrow \tan 2 \alpha\left[\frac{\tan 2 \beta+\tan 2 \gamma}{1+\tan 2 \beta \tan 2 \gamma}\right]=1 \\
& \Rightarrow \quad \tan 2 \alpha \cdot \tan (2 \beta+2 \gamma)=1 \\
& \Rightarrow \quad \tan (2 \beta+2 \gamma)=\cot 2 \alpha \\
& \Rightarrow \quad \tan \{30+36\}=\cot 24 \\
& \Rightarrow \quad \tan 66^{\circ}=\cot 24^{\circ} \\
& \Rightarrow \quad \tan 66^{\circ}=\tan \left(90^{\circ}-66^{\circ}\right) \\
& \Rightarrow \tan 66^{\circ}=\tan 66^{\circ} \text { which is true } \\
&
\end{aligned}
$$
$$
\begin{aligned}
\mathrm{R} \rightarrow & \tan \frac{A}{2} \tan \frac{B}{2}+\tan \frac{B}{2} \tan \frac{C}{2}=1-\tan \frac{C}{2} \tan \frac{A}{2} \\
& \frac{\tan \frac{B}{2}\left[\tan \frac{A}{2}+\tan \frac{C}{2}\right]}{1-\tan \frac{A}{2} \tan \frac{C}{2}}=1 \\
\Rightarrow & \tan \left[\frac{A}{2}+\frac{C}{2}\right]=\cot \frac{B}{2} \Rightarrow A+B+C=180 \\
\Rightarrow & \quad A+C=180-B \Rightarrow \frac{A+C}{2}=90-\frac{B}{2} \\
\Rightarrow & \tan \left[90^{\circ}-\frac{B}{2}\right]=\cot \frac{B}{2} \Rightarrow \cot \frac{B}{2}=\cot \frac{B}{2}
\end{aligned}
$$
$\mathrm{R}$ is true and it is correct explanation of $A$.
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