If f(t) = frac{1 + operatorname{cosec} t}{1 - | vedclass

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(B) Given,$f(t) = \frac{1 + \operatorname{cosec} t}{1 - \operatorname{cosec} t}$.Taking the natural logarithm on both sides,$\ln f(t) = \ln(1 + \opera

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$4 \operatorname{cosec} 2t$

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(B) Given,$f(t) = \frac{1 + \operatorname{cosec} t}{1 - \operatorname{cosec} t}$.Taking the natural logarithm on both sides,$\ln f(t) = \ln(1 + \operatorname{cosec} t) - \ln(1 - \operatorname{cosec} t)$.Differentiating with respect to $t$:$\frac{f^{\prime}(t)}{f(t)} = \frac{-\operatorname{cosec} t \cot t}{1 + \operatorname{cosec} t} - \frac{\operatorname{cosec} t \cot t}{1 - \operatorname{cosec} t}$.$\frac{f^{\prime}(t)}{f(t)} = -\operatorname{cosec} t \cot t \left( \frac{1}{1 + \operatorname{cosec} t} + \frac{1}{1 - \operatorname{cosec} t} \right)$.$\frac{f^{\prime}(t)}{f(t)} = -\operatorname{cosec} t \cot t \left( \frac{1 - \operatorname{cosec} t + 1 + \operatorname{cosec} t}{1 - \operatorname{cosec}^2 t} \right)$.$\frac{f^{\prime}(t)}{f(t)} = -\operatorname{cosec} t \cot t \left( \frac{2}{-\cot^2 t} \right) = \frac{2 \operatorname{cosec} t \cot t}{\cot^2 t} = \frac{2 \operatorname{cosec} t}{\cot t} = 2 \sec t \operatorname{cosec} t$.Since $2 \sec t \operatorname{cosec} t = \frac{2}{\sin t \cos t} = \frac{4}{2 \sin t \cos t} = \frac{4}{\sin 2t} = 4 \operatorname{cosec} 2t$.Thus,$g(t) = 4 \operatorname{cosec} 2t$.

If $f(t) = \frac{1 + \operatorname{cosec} t}{1 - \operatorname{cosec} t}$ for $0 < t < \frac{\pi}{2}$ and $f^{\prime}(t) = f(t) g(t)$,then $g(t) =$

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