If \(f(t)=\frac{1+\operatorname{cosec} t}{1-\operatorname{cosec} t}\) for \(0 < t < \frac{\pi}{2}\) and…

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 \(g(t)=\)
  1. \(-4 \operatorname{cosec} 2 t\)
  2. \(4 \operatorname{cosec} 2 t\)
  3. \(2 \sin 2 t\)
  4. \(4 \operatorname{cosec} t\)

Solution

Given, \(f(t)=\frac{1+\operatorname{cosec} t}{1-\operatorname{cosec} t}\) Differentiating w.r.t. \(t\), we get \((1-\operatorname{cosec} t)(-\operatorname{cosec} t \cot t)\) \(\begin{aligned} f^{\prime}(t) & =\frac{-(1-\operatorname{cosec} t)(\operatorname{cosec} t \cot t)}{(1-\operatorname{cosec} t)^2} \\ & =\frac{-\operatorname{cosec} t \cot t+\operatorname{cosec}^2 t \cot t}{(1-\operatorname{cosec} t)^2} \\ & =\frac{-2 \operatorname{cosec} t \cot t}{(1-\operatorname{cosec} t)^2} \times \frac{1+\operatorname{cosec} t}{1+\operatorname{cosec} t} \\ & =\left(\frac{1+\operatorname{cosec} t}{1-\operatorname{cosec} t}\right)\left(-\frac{2 \operatorname{cosec} t \cot t}{1-\operatorname{cosec}^2 t}\right) \\ & =f(t) \frac{2 \operatorname{cosec} t \cot t}{\cot ^2 t}=f(t) \frac{2 \operatorname{cosec}^2 t}{\cot ^2 t} \\ & =f(t) \frac{2}{\sin t \cos t}=f(t)(4 \operatorname{cosec} 2 t)=f(t) g(t) \quad \text{[given]} \end{aligned}\) So, \(g(t)=4 \operatorname{cosec} 2 t\)

Asked in: AP EAMCET 2019 (22 Apr Shift 1)

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