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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Question : If $6 \sec \theta=10$, then find the value of $\frac{5 \operatorname{cosec} \theta-3 \cot \theta}{4 \cos \theta+3 \sin \theta}$.
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Question : If $0^{\circ}< \theta< 90^{\circ}$ and $\operatorname{cosec \theta} =\cot^{2}\theta$, then the value of expression $\operatorname{cosec^{4}\theta}–\operatorname{2cosec^{2}\theta}-\cot^{2}\theta$ is equal to:
Question : What is the value of $\sqrt{\frac{\operatorname{cosec} A+1}{\operatorname{cosec} A-1}}+\sqrt{\frac{\operatorname{cosec} A-1}{\operatorname{cosec} A+1}}$?
Question : The value of $\frac{1}{\sin \theta}-\frac{\cot ^2 \theta}{1+\operatorname{cosec} \theta}$ is:
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