Question : What is the value of $\frac{\cot \theta+\operatorname{cosec} \theta-1}{\cot \theta-\operatorname{cosec} \theta+1}$?
Option 1: $2 \sec \theta$
Option 2: $2 \operatorname{cosec} \theta$
Option 3: $2 \cot \theta$
Option 4: $\operatorname{cosec} \theta+\cot \theta$
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Correct Answer: $\operatorname{cosec} \theta+\cot \theta$
Solution :
Given, $\frac{\cot \theta+\operatorname{cosec} \theta-1}{\cot \theta-\operatorname{cosec} \theta+1}$
We know, $\operatorname{cosec^2}θ−\cot^2θ=1$
$=\frac{\cot \theta+\operatorname{cosec} \theta-(\operatorname{cosec^2}θ−\cot^2θ)}{\cot \theta-\operatorname{cosec} \theta+1}$
$=\frac{\cot \theta+\operatorname{cosec} \theta-(\operatorname{cosec}θ−\cotθ)(\operatorname{cosec}θ+\cotθ)}{\cot \theta-\operatorname{cosec} \theta+1}$
$=\frac{(\operatorname{cosec}θ+\cotθ)[1-(\operatorname{cosec}θ-\cotθ)]}{\cot \theta-\operatorname{cosec} \theta+1}$
$=\frac{(\operatorname{cosec}θ+\cotθ)[1-\operatorname{cosec}θ+\cotθ)]}{\cot \theta-\operatorname{cosec} \theta+1}$
$=\operatorname{cosec}θ+\cotθ$
Hence, the correct answer is $\operatorname{cosec}θ+\cotθ$.
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