Question 741485
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*[tex \LARGE \tan\varphi] is the reciprocal of *[tex \LARGE \cot\varphi]


*[tex \LARGE \cot\varphi\ =\ \frac{\cos\varphi}{\sin\varphi}], so  *[tex \LARGE \cos\varphi\ =\ 7\sin\varphi] if *[tex \LARGE \cot\varphi\ =\ 7]


Then *[tex \LARGE \cos^2\varphi\ =\ 49\sin^2\varphi]


But since *[tex \LARGE \cos^2\psi\ +\ \sin^2\psi\ =\ 1], we can write


*[tex \LARGE \cos^2\varphi\ =\ 49\ -\ 49\cos^2\varphi]  which is to say *[tex \LARGE \cos^2\varphi\ =\ \frac{49}{50}] from which we get *[tex \LARGE \cos\varphi\ =\ \pm\frac{7\sqrt{2}}{10}]


Then since we know that *[tex \LARGE \cos\varphi\ =\ 7\sin\varphi], it is clear that *[tex \LARGE \sin\varphi\ =\ \pm\frac{\sqrt{2}}{10}]


Now that you have *[tex \LARGE \sin], calculate *[tex \LARGE \csc\varphi] by noting that *[tex \LARGE \sin] is the reciprocal of *[tex \LARGE \csc]


Likewise, *[tex \LARGE \cos] is the reciprocal of *[tex \LARGE \sec]


Don't forget to rationalize your denominators in all casess.


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
<font face="Math1" size="+2">Egw to Beta kai to Sigma</font>
My calculator said it, I believe it, that settles it
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