Trigonometry . help

I will do the second one may be?

\(sinA(1+cot^2A) = sinA+\frac{sinA}{tan^2A}=sinA+\frac{sinAcos^2A}{sin^2A}=\frac{sin^2A}{sinA}+\frac{cos^2A}{sinA}=\frac{1}{sinA}=cscA \)

Hi Cath,

I'll do the first one, you  try each one yourself before you post it.  

1)

\(\frac{tanxCosec^2x}{1+tan^2x}\\ =(tanxCosec^2x)\;\div (1+tan^2x)\\ =(\frac{sinx}{cosx} \frac{1}{sin^2x})\;\div (\frac{cos^2x}{cos^2x}+\frac{sin^2x}{cos^2x})\\ =\frac{1}{cosxsinx} \;\div \frac{1}{cos^2x}\\ =\frac{1}{cosxsinx} \;\times \frac{cos^2x}{1}\\ =\frac{1}{sinx} \;\times \frac{cosx}{1}\\ =cot(x)\)

I now have time to do some of the remaining questions :D good news.

3)\(\frac{siny+tany+cosy}{tany}=siny\times coty+1+cosy\times coty=cosy+cosy\times secy+cosy\times coty=cosy(1+secy+coty)\)

4)\(cscUsecU-cotU=\frac{1}{sinUcosU}-\frac{1}{tanU}=\frac{tanU-sinUcosU}{sinUcosUtanU}=\frac{sinU(1-cos^2U)}{sin^2UcosU}=\frac{sin^3U}{sin^2UcosU}=\frac{sinU}{cosU}=tanU\)

5)\(\frac{tanA+1}{sec^4A}=\frac{\frac{sinA+cosA}{cosA}}{\frac{1}{cos^4A}}=cos^3AsinA+cos^4A=cos^3A(sinA+cosA)\)