If f(t)=1+cosec⁡t1−cosec⁡t for 0<t<π2 and f′(t)=f(t)g(t), then g(t)=

f(t)=1+cosec⁡t1−cosec⁡t=sin⁡t+1sin⁡t−1⇒f′(t)=(sin⁡t−1)cos⁡t−(sin⁡t+1)cos⁡t(sin⁡t−1)2=cos⁡t(sin⁡t−1−sin⁡t−1)(sin⁡t−1)2=−2cos⁡t(sin⁡t−1)2 =(sin⁡t+1)(sin⁡t−1)×(−2cos⁡t)(sin⁡t−1)(sin⁡t+1)=f(t)−2cos⁡tsin2⁡t−1=f(t)2cos⁡tcos2⁡t=f(t)(2sec⁡t)∴ g(t)=2sec⁡t

If f(t)=1+cosec⁡t1−cosec⁡t for 0<t<π2 and f′(t)=f(t)g(t), then g(t)=

f(t)=1+cosec⁡t1−cosec⁡t=sin⁡t+1sin⁡t−1⇒f′(t)=(sin⁡t−1)cos⁡t−(sin⁡t+1)cos⁡t(sin⁡t−1)2=cos⁡t(sin⁡t−1−sin⁡t−1)(sin⁡t−1)2=−2cos⁡t(sin⁡t−1)2 =(sin⁡t+1)(sin⁡t−1)×(−2cos⁡t)(sin⁡t−1)(sin⁡t+1)=f(t)−2cos⁡tsin2⁡t−1=f(t)2cos⁡tcos2⁡t=f(t)(2sec⁡t)∴ g(t)=2sec⁡t

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If $f (t) = \frac{1 + cosec t}{1 - cosec t} for 0 < t < \frac{π}{2} and f^{'} (t) = f (t) g (t), then g (t) =$

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$2 \sec t$

$4 cosec 2 t$

$2 \sin 2 t$

$4 cosec t$

answer is A.

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Detailed Solution

$\begin{array}{l} f (t) = \frac{1 + cosec t}{1 - cosec t} = \frac{\sin t + 1}{\sin t - 1} \\ \Rightarrow f^{'} (t) = \frac{(\sin t - 1) \cos t - (\sin t + 1) \cos t}{(\sin t - 1)^{2}} = \frac{\cos t (\sin t - 1 - \sin t - 1)}{(\sin t - 1)^{2}}) = \frac{- 2 \cos t}{(\sin t - 1)^{2}} \end{array}$

$\begin{array}{l} = \frac{(\sin t + 1)}{(\sin t - 1)} \times \frac{(- 2 \cos t)}{(\sin t - 1) (\sin t + 1)} = f (t) (\frac{- 2 \cos t}{\sin^{2} t - 1}) = f (t) \frac{2 \cos t}{\cos^{2} t} = f (t) (2 \sec t) \\ ∴ g (t) = 2 \sec t \end{array}$