Question

I am trying to prove that (tanx+cotx)/(secx+cscx) is equal to 1/(cosx+sinx) ... can someone show me the steps?

Answer 1

anyone?

Answer 2

Just write all the funny expression out as sin/cos and do the algebra, it will just fall out...

Answer 3

right, i'm at ((sinx/cosx)+(cosx/sinx))/((1/cosx)+(1/sinx)) and as usual the basic algebra is messing me up as to how i "flip" the fractions.....

Answer 4

((1/cosx)+(1/sinx)) do the addition first and then turn it upside down n multiply.

Answer 5

Rewrite as:
\[\huge\frac{\frac{\sin(x)}{\cos(x)}+\frac{\cos(x)}{\sin(x)}}{\frac{1}{\cos(x)}+\frac{1}{\sin(x)}}\]
No get a common denominator in the top and bottom.
\[\huge\frac{\frac{\sin^2(x)+\cos^2(x)}{\sin(x)\cos(x)}}{\frac{\sin(x)+\cos(x)}{\sin(x)\cos(x)}}\]
Well: sin^2(x)+cos^2(x)=1
So you have:
\[\huge\frac{\frac{1}{\sin(x)\cos(x)}}{\frac{\sin(x)+\cos(x)}{\sin(x)\cos(x)}}=\frac{\sin(x)\cos(x)}{\sin(x)\cos(x)(\sin(x)+\cos(x))}\]
\[\huge=\frac{1}{\sin(x)+\cos(x)}\]

Answer 6

hory smokes. ty ty

Answer 7

I made it big so you could read the fractions, they were kinda jumbled. :P

Answer 8

you are mighty in the ways of the trig

Answer 9

Haha, I try :P I've just seen it alot so the approaches come easy. The approach is the most important part. The algebra can be done by computers :P