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OpenStudy (anonymous):
simplify: 1/cot(x)-1/2sinxsecx
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OpenStudy (anonymous):
algebra is easier i you replace \(\cos(x)\) by \(a\) and \(\sin(x)\) by \(b\)
OpenStudy (anonymous):
\[\frac{1}{\frac{a}{b}}-\frac{1}{2b\times \frac{1}{a}}\]
OpenStudy (anonymous):
you get
\[\frac{b}{a}-\frac{a}{2b}\]
OpenStudy (anonymous):
this isn't algebra it's trigonometry. I don't understand what you are doing.
OpenStudy (anonymous):
then subtract, get \[\frac{2b^2-a^2}{2ab}\]
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OpenStudy (anonymous):
it is algebra, just with trig functions
OpenStudy (anonymous):
is the question
\[\frac{1}{\cot(x)}-\frac{1}{2\sin(x)\sec(x)}\]?
OpenStudy (anonymous):
yes, but i need to answer it using trig identies.
OpenStudy (anonymous):
the reciprocal of cotangent is tangent, and the reciprocal of secant is cosine
OpenStudy (anonymous):
so you can write
\[\tan(x)-\frac{\cos(x)}{2\sin(x)}\] if you like
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OpenStudy (anonymous):
is the answer tanx
OpenStudy (anonymous):
no
OpenStudy (anonymous):
you can subtract if you like
OpenStudy (anonymous):
but the answer is not some simple trig function, at least not to the question you posted
there is a 2 in the denominator right?
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