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If $\mathrm{p}=\operatorname{cosec} \theta-\cot \theta$ and $\mathrm{q}=(\operatorname{cosec} \theta+\cot \theta)^{-1}$, then which
one of the following is correct?
Options:
one of the following is correct?
Solution:
1184 Upvotes
Verified Answer
The correct answer is:
$\mathrm{p}=\mathrm{q}$
& \mathrm{p}=\operatorname{cosec} \theta-\cot \theta \\
& \mathrm{q}=(\operatorname{cosec} \theta+\cot \theta)^{-1} \\
& \Rightarrow \frac{1}{q}=\operatorname{cosec} \theta+\cot \theta \\
& \text { We know, } \operatorname{cosec}^{2} \theta-\cot ^{2} \theta=1 \\
& \Rightarrow(\operatorname{cosec} \theta+\cot \theta)(\operatorname{cosec} \theta-\cot \theta)=1 \\
& \Rightarrow\left(\frac{1}{q}\right)(p)=1 \\
& \Rightarrow \mathrm{p}=\mathrm{q}
\end{aligned}
$$
& \mathrm{q}=(\operatorname{cosec} \theta+\cot \theta)^{-1} \\
& \Rightarrow \frac{1}{q}=\operatorname{cosec} \theta+\cot \theta \\
& \text { We know, } \operatorname{cosec}^{2} \theta-\cot ^{2} \theta=1 \\
& \Rightarrow(\operatorname{cosec} \theta+\cot \theta)(\operatorname{cosec} \theta-\cot \theta)=1 \\
& \Rightarrow\left(\frac{1}{q}\right)(p)=1 \\
& \Rightarrow \mathrm{p}=\mathrm{q}
\end{aligned}
$$
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