<PRE><B>Question 82694</B>
Hi,&lt;br /&gt;

i&#39;m very sorry if I&#39;m wrong, but I get the impression you just want the answers, so I&#39;m going to give some helpful advice instead. If you have tried it then please post where you&#39;re stuck.&lt;br /&gt;

&lt;b&gt;First question:&lt;/b&gt; *[tex \operatorname{cosec}(x)\tan(-x)] For questions like this, I always recommend you go back to basics. Replace everything in terms of sine and cosine. Well, cosec=1/sin and tan=sin/cos. (You need to remember these!) I&#39;m sure you can take it from there. Don&#39;t forget what sin(-x) and cos(-x) are.&lt;br /&gt;

&lt;b&gt;Second Question:&lt;/b&gt; *[tex \cot(x)\cos(\frac{pi}{2}-x)] The idea is the same as last time, write everything in terms of sines and cosines. If you don&#39;t know cot=cos/sin. However this time we need to get rid of that pi/2. See what you can do with the addition formula for cosine (you should also know this) *[tex \cos(A+B)=\cos(A)\cos(B)-\sin(A)\sin(B)].&lt;br /&gt;

&lt;b&gt;Third Question:&lt;/b&gt; I&#39;m sure there are many ways to do this. My method is to write everything in terms of cos(2x). What do you get if you let A=B in the addition formula given above for cosine. can you combine this with the identity *[tex sin^2 + cos^2 = 1] to get *[tex sin^2(x)] as a function of *[tex \cos(2x)]. let me know how you get on with this one because it&#39;s quite hard. If you get stuck post back with your working.&lt;br /&gt;

Hope that helps,
Kev</PRE>