Question #352ac

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Question #352ac

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Answer 1 · 2018-02-16T16:34:37

$tan (θ)$ $tan (theta)$

Explanation:

$sec (θ) \equiv \frac{1}{cos (θ)}$ $sec(theta) -= 1/cos(theta)$

$\Rightarrow sin (θ) sec (θ) \equiv sin (θ) \cdot \frac{1}{cos (θ)}$ $=> sin(theta)sec(theta) -= sin(theta)*1/cos(theta)$

$= \frac{sin (θ)}{cos (θ)}$ $=sin(theta)/cos(theta)$

$= tan (θ)$ $=tan(theta)$

Answer 2 · 2018-02-16T16:37:56

$tan (θ) = sin (θ) sec (θ)$ $tan(theta)=sin(theta)sec(theta)$

Explanation:

$sec (θ)$ $sec(theta)$ is another way of writing $\frac{1}{cos (θ)}$ $1/cos(theta)$

So $sin (θ) \times sec (θ) = sin (θ) \times \frac{1}{cos (θ)} = \frac{sin (θ)}{cos (θ)}$ $sin(theta)xxsec(theta) = sin(theta)xx1/cos(theta) = sin(theta)/cos(theta)$

but we also have $tan (θ) = \frac{sin (θ)}{cos (θ)}$ $tan(theta)=sin(theta)/cos(theta)$

Thus $tan (θ) = sin (θ) sec (θ)$ $tan(theta)=sin(theta)sec(theta)$

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Question #352ac

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