Assertion (A):If $\mathrm{A}=10^{\circ}, \mathrm{B}=16^{\circ}, \mathrm{C}=19^{\circ}$, then $\tan 2…
Assertion (A):If $\mathrm{A}=10^{\circ}, \mathrm{B}=16^{\circ}, \mathrm{C}=19^{\circ}$, then $\tan 2 \mathrm{~A}$ $\tan 2 \mathrm{~B}+\tan 2 \mathrm{~B} \tan 2 \mathrm{C}+\tan 2 \mathrm{C} \tan 2 \mathrm{~A}=1$.
Reason (R): If $A+B+C=180^{\circ},=\cot \frac{A}{2} \cdot \cot \frac{B}{2} \cdot \cot \frac{C}{2}$ $=\cot \frac{\mathrm{A}}{2} \cdot \cot \frac{\mathrm{~B}}{2} \cdot \cot \frac{\mathrm{C}}{2}$
Which of the following is correct?
- Both (A) and (R) are true and (R) is the correct explanation of (A)
- Both (A) and (R) are true and (R) is NOT correct explanation of (A)
- (A) is true, (R) is false
- (A) is false, (R) is true
Solution
Given that $A=10^{\circ}, B=16^{\circ}$ and $C=19^{\circ}$
$A+B+C=45^{\circ} \Rightarrow 2 A+2 B+2 C=90^{\circ}$...(i)
We know that, $\tan (2 A+2 B+2 C)$
$\begin{aligned}
& =\frac{\tan 2 A+\tan 2 B+\tan 2 C-\tan 2 A \tan 2 B \tan 2 C}{1-\tan 2 A \cdot \tan 2 B-\tan 2 B \cdot \tan 2 C-\tan 2 C \cdot \tan 2 A} \\
& \infty=\frac{\tan 2 A+\tan 2 B+\tan 2 C-\tan 2 A \tan 2 B \tan 2 C}{1-\tan 2 A \cdot \tan 2 B-\tan 2 B \cdot \tan 2 C-\tan 2 C \cdot \tan 2 A} \\
& \Rightarrow \tan 2 A \cdot \tan 2 B+\tan 2 B \cdot \tan 2 C+\tan 2 C \cdot \tan 2 A=1
\end{aligned}$
So, Assertion is correct.
Now divide by $\tan 2 A \cdot \tan 2 B \cdot \tan 2 C$ both side
$\begin{aligned}
& \frac{1}{\tan 2 A}+\frac{1}{\tan 2 B}+\frac{1}{\tan 2 C}=\frac{1}{\tan 2 A \cdot \tan 2 B \cdot \tan 2 C} \\
& \Rightarrow \cot 2 A+\cot 2 B+\cot 2 C=\cot 2 A \cdot \cot 2 B \cdot \cot 2 C \\
& \Rightarrow \cot \frac{A}{2}+\cot \frac{B}{2}+\cot \frac{C}{2}=\cot \frac{A}{2} \cdot \cot \frac{B}{2} \cdot \cot \frac{C}{2}
\end{aligned}$
Also $A+B+C=180^{\circ}$ [from (i)]
So, Reason is correct.
Asked in: AP EAMCET 2024 (21 May Shift 2)
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