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prove the following identities csc(theta) - sin (theta) = cos(theta*cot(theta)
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probably nothing but a bunch of algebra. easier to do if you replace cosine by "a" and sine by "b" and compute \[\frac{1}{b}-b=a\times \frac{a}{b}\]
on the left hand side you get \[\frac{1-b^2}{b}\] on the right hand side you get \[\frac{a^2}{b}\] so you want so show that \[\frac{1-b^2}{b}=\frac{a^2}{b}\] now go back to trig and get \[\frac{1-\sin^2(x)}{\sin(x)}=\frac{\cos^2(x)}{\sin(x)}\] and since \[1-\sin^2(x)=\cos^2(x)\] you get what you want
thank you, can i ask one more question?
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