Question: "Derivatives involving other trigonometric functions Find the derivative..."

Michael Dennis

Answered question

2021-11-24

Derivatives involving other trigonometric functions Find the derivative of the following functions.

y = \frac{\tan t}{1 + \sec t}

Answer & Explanation

Michele Grimsley

Beginner2021-11-25Added 19 answers

To derive the other trigonometric function,
$y = \frac{\tan t}{1 + \sec t}$
$= \frac{\frac{\sin t}{\cos t}}{1 + \frac{1}{\cos t}}$
$= \frac{\frac{\sin t}{\cos t}}{\frac{\cos t + 1}{\cos t}}$
Since, $\frac{\sin t}{\cos t} = \tan t, \sec t = \frac{1}{\cos t}$
$y = \frac{\sin t}{1 + \cos t}$
Differentiate the equation (1) with respect to “x”
Apply quotient rules of derivatives,
$y' \frac{d}{d t} (\begin{matrix} \frac{\sin t}{1 + \cos t} \end{matrix})$
$= \frac{(1 + \cos t) \frac{d}{d t} \sin t - \sin t \frac{d}{d t} (1 + \cos t)}{{(1 + \cos t)}^{2}}$
$= \frac{(1 + \cos t) \cos t - \sin t (0 - \sin t)}{{(1 + \cos t)}^{2}}$
$= \frac{\cos t + \cos^{2} t + \sin^{2} t}{{(1 + \cos t)}^{2}}$
Further solving it,
$= \frac{\cos t + 1}{{(1 + \cos t)}^{2}}$ Since,
$\sin^{2} t + \cos^{2} t = 1$
$= \frac{1 + \cos t + 1}{{(1 + \cos t)}^{2}}$
$y' = \frac{1}{1 + \cos t}$

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