Question 283917
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Almost right, but not quite.  You say that the left hand side is equal to the right hand side.  In fact the two sides are <b><i>identical</i></b>, and you should have asked for someone to prove the identity.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \sec\phi\ \equiv\ \frac{1}{cos\phi}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \tan\phi\ \equiv\ \frac{sin\phi}{cos\phi}]


So your statement becomes:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{1}{cos{x}}\ -\ \frac{sin^2{x}}{cos{x}}\ \equiv\ \cos{x}]


Multiplying both sides by *[tex \Large \cos{x}]:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 1\ -\ sin^2{x}\ \equiv\ \cos^2{x}]


Which is to say


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \cos^2{x}\ +\ \sin^2{x}\ \equiv\ 1]


for which you should already have proved the theorem.


John
*[tex \LARGE e^{i\pi} + 1 = 0]
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