I can do pretty much all the proofs except these two, though i've tried to make a start. Note that in the questions the angle is given as theta, but for the sake of my keyboard i've changed it to angle 'x'. Q15: Prove: Tan2xsecx = 2sinxsec2x. So I need to show the two sides are equal. First i worked on the left hand side. Using the double angle conversion formulae it becomes: =2sinxsec2x =(2tanx)/(1-tan^2 x)secx =(2tanx)/(1-tan^2 x)1/cosx =(2tanx)/(1-tan^2 x)cosx =(2tanx)/(cosx-tan^2cosx) I find it difficult to find exactly the right form to convert back to double angles via manipulation so i converted the right hand side to single angles (to see what form i needed it in). =2sinxsec2x =2sinx multiplied by 1/(cos2x) =2sinx multipled by 1/(cos^2x - sin^2x) =2sinx/(cos^2x - sin^2x) So far i have: =(2tanx)/(cosx-tan^2cosx) = 2sinx/(cos^2x - sin^2x) Is that correct so far and how can i go on to show they are exactly the same? And the other one: Q17: Prove that: 1/(cosx-sinx) - 1/(cosx+sinx) = 2sinxsec2x The LHS consisted entirely of single angles so i left that for the time being and worked on the RHS. =2sinxsec2x =2sinx multiplied by 1/(cos2x) =2sinx multiplied by 1/(cos^2x - sin^2x) =2sinx/(cos^2x - sin^2x) =2sinx/(1-sin^2x-sin^2x) =2sinx/(1-2sin^x) And i have no idea where to go from there. Any help on either of the questions would be much appreciated.