How can you prove: secθ = cscθtanθ

1 answer:

3 0

$\sec \theta=\csc \theta\tan \theta\\\\ \dfrac{1}{\cos \theta}=\dfrac{1}{\sin \theta}\cdot\dfrac{\sin \theta}{\cos \theta}\\\\ \dfrac{1}{\cos \theta}=\dfrac{1}{\cos \theta}$

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Step-by-step explanation:

See attachment for the figure.

<u>For D'</u>, rotating the point (2, 2) 90° counterclockwise. This maps each point (x, y)→(-y, x); it takes (2, 2)→(-2, 2).

For C'', rotating the point (1, 3) 90° clockwise. This maps each point (x, y)→(y, -x); it takes (1, 3)→(3, -1).

For A''', rotating the point (-4, 2) 180° clockwise. This maps every point (x, y)→(-x, -y); it takes (-4, 2)→(4, -2).

For B'', rotating the point (-2, 4) 270° counterclockwise. This is the similar as rotating the point 90° clockwise. This maps every point (x, y)→(y, -x); it takes (-2, 4)→(4, 2).