What is the tangent of 1/16 tan^2(x)+cos(x) ?

The tangent of 1/16 tan^2(x)+cos(x) is

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tangent of 1/16 tan^2(x)+cos(x)

tangent of

\frac{1}{16} tan^{2} (x) + cos (x)

y = (\frac{1}{8} sec^{2} (a_{0}) tan (a_{0}) - sin (a_{0})) x + \frac{1}{16} tan^{2} (a_{0}) + cos (a_{0}) - (\frac{1}{8} sec^{2} (a_{0}) tan (a_{0}) - sin (a_{0})) a_{0}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, \frac{1}{16} tan^{2} (a_{0}) + cos (a_{0}))

Find the slope of

y = \frac{1}{16} tan^{2} (x) + cos (x) : \frac{d y}{d x} = \frac{1}{8} sec^{2} (x) tan (x) - sin (x)

Compute the slope of

y = \frac{1}{16} tan^{2} (x) + cos (x)

x = a_{0} : m = \frac{1}{8} sec^{2} (a_{0}) tan (a_{0}) - sin (a_{0})

Find the line with slope m=

\frac{1}{8} sec^{2} (a_{0}) tan (a_{0}) - sin (a_{0})

and passing through

(a_{0}, \frac{1}{16} tan^{2} (a_{0}) + cos (a_{0})) : y = (\frac{1}{8} sec^{2} (a_{0}) tan (a_{0}) - sin (a_{0})) x + \frac{1}{16} tan^{2} (a_{0}) + cos (a_{0}) - (\frac{1}{8} sec^{2} (a_{0}) tan (a_{0}) - sin (a_{0})) a_{0}

y = (\frac{1}{8} sec^{2} (a_{0}) tan (a_{0}) - sin (a_{0})) x + \frac{1}{16} tan^{2} (a_{0}) + cos (a_{0}) - (\frac{1}{8} sec^{2} (a_{0}) tan (a_{0}) - sin (a_{0})) a_{0}

x \to 1 + lim (3 x)

\int \frac{x ^{2} + 10 x - 1}{x ^{3} - x} d x

\int - 2 y 2 - y^{2} d y

\int_{- \frac{π}{3}}^{0} 2 sec^{3} (x) d x

What is the tangent of 1/16 tan^2(x)+cos(x) ?
The tangent of 1/16 tan^2(x)+cos(x) is