Can you help me solve this trig identity question?

[x.xmedzx.x]

hey guyz...ok iv been trying to figure this question out for so long...and i jus can't.i get up to a certain point and then i jus get confused.so if anyone can help me that would be great!

Prov that:
Cos^2x + Cotx ÷ Cos^2x – Cotx = Cos^2x (tanx) + 1 ÷ Cos^2x (tanx) -1

 

[courtrigrad]

$$\frac{\cos^{2}x + \cot x}{\cos^{2}x-\cot x} = \frac{\cos^{2}x(\tan x)+1}{\cos^{2} x(\tan x) -1}$$

Convert the left side to sines and cosines:

$$\frac{\cos^{2}x + \frac{\cos x}{\sin x}}{\cos^{2}x - \frac{\cos x}{\sin x}} = \frac{\cos^{2}x\sin x + \cos x}{\sin x}\frac{\sin x}{\cos^{2}x\sin x - \cos x} = \frac{\cos^{2}x\sin x + \cos x}{\cos^{2}x\sin x - \cos x} = \frac{\cos^{2}x(\sin x + \frac{1}{\cos x})}{\cos^{2}x(\sin x - \frac{1}{\cos x})}$$.

Can you go from there?

 

[x.xmedzx.x]

umm ok the first setp i got...but the 2nd one I am a little bit confused as to what you did...

 

[Hootenanny]

umm ok the first setp i got...but the 2nd one I am a little bit confused as to what you did...

Note that;

$$\cot\theta=\frac{1}{\tan\theta}=\frac{1}{\frac{\sin\theta}{\cos\theta}}=\frac{\cos\theta}{\sin\theta}$$