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If $\mathrm{p}=\tan \left(-\frac{11 \pi}{6}\right), \mathrm{q}=\tan \left(\frac{21 \pi}{4}\right)$ and $\mathrm{r}=\cot \left(\frac{283 \pi}{6}\right)$,
then which of the following is/are correct?
1. The value of $p \times r$ is 2 .
2. $\mathrm{p}, \mathrm{q}$ andr are in G.P. Select the correct answer using the code given below:
Options:
then which of the following is/are correct?
1. The value of $p \times r$ is 2 .
2. $\mathrm{p}, \mathrm{q}$ andr are in G.P. Select the correct answer using the code given below:
Solution:
1160 Upvotes
Verified Answer
The correct answer is:
2 only
\begin{array}{l}
\mathrm{p}=\tan \left(-\frac{11 \pi}{6}\right) \\
\mathrm{p}=-\tan \left(2 \pi-\frac{\pi}{6}\right) \\
\mathrm{p}=\tan \frac{\pi}{6}=\frac{1}{\sqrt{3}} \\
\mathrm{q}=\tan \left(\frac{21 \pi}{4}\right)=\tan \left(6 \pi-\frac{3 \pi}{4}\right) \\
\mathrm{q}=-\tan \frac{3 \pi}{4}=+\tan \frac{\pi}{4}=1 \\
\mathrm{r}=\cot \left(\frac{283 \pi}{6}\right)=\cot \left(46 \pi+\frac{7 \pi}{6}\right)=\cot \left(\pi+\frac{\pi}{6}\right) \\
\mathrm{r}=\cot \frac{\pi}{6}=\sqrt{3} \\
\mathrm{p} \times \mathrm{r}=\frac{1}{\sqrt{3}} \times \sqrt{3}=1 \\
\therefore \text { Statement(1) is incorrect. }
\end{array}
also
$\frac{\mathrm{p}}{\mathrm{q}}=\frac{\mathrm{q}}{\mathrm{r}}=\frac{1}{\sqrt{3}}$
So p, q, rare in G.P. $\therefore$ Statement $(2)$ is correct.
\mathrm{p}=\tan \left(-\frac{11 \pi}{6}\right) \\
\mathrm{p}=-\tan \left(2 \pi-\frac{\pi}{6}\right) \\
\mathrm{p}=\tan \frac{\pi}{6}=\frac{1}{\sqrt{3}} \\
\mathrm{q}=\tan \left(\frac{21 \pi}{4}\right)=\tan \left(6 \pi-\frac{3 \pi}{4}\right) \\
\mathrm{q}=-\tan \frac{3 \pi}{4}=+\tan \frac{\pi}{4}=1 \\
\mathrm{r}=\cot \left(\frac{283 \pi}{6}\right)=\cot \left(46 \pi+\frac{7 \pi}{6}\right)=\cot \left(\pi+\frac{\pi}{6}\right) \\
\mathrm{r}=\cot \frac{\pi}{6}=\sqrt{3} \\
\mathrm{p} \times \mathrm{r}=\frac{1}{\sqrt{3}} \times \sqrt{3}=1 \\
\therefore \text { Statement(1) is incorrect. }
\end{array}
also
$\frac{\mathrm{p}}{\mathrm{q}}=\frac{\mathrm{q}}{\mathrm{r}}=\frac{1}{\sqrt{3}}$
So p, q, rare in G.P. $\therefore$ Statement $(2)$ is correct.
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