Trig Question (1 Viewer)

[s2Vicki]

Can anyone help me this question?

Simplify (1+cot x / cosec x) - (sec x / tan x + cot x)

 

[Drongoski]

 

[Gussy Booo]

Hey there ! Ok here we go

So lets work on the left hand side of this question.

(1+cot /cosecx) - (1 + cosx/sinx) / (1/sinx)

Add (1 + cosx/sinx)

= (sinx+cosx/sinx) / (1/sinx)

Sin x is the denominator in both

.'. Sinx+cosx

Thats the left hand side.

Lets work with the right

(secx/tanx+cot)

= (1/cosx) / (sinx/cosx)/ (cosx/sinx)

= (1/cosx) / (sin^2x+cos^2x / sinxcosx)

sin^2x+cos^2x = 1

therefore: (1/cosx) / (1/ sinxcosx)

solce that and you get Sinx

Now we bring out last result and we get


Sinx+cosx-sinx

.'. = cosx

 

[s2Vicki]

Add (1 + cosx/sinx)

= (sinx+cosx/sinx) / (1/sinx)

Thanks so much. But I don't get a couple of lines.
Why do you add (1 + cosx/sinx) and how do you get (sinx+cosx/sinx) / (1/sinx) from that?

 

[gurmies]

Look at drongoski's solution =]

 

[s2Vicki]

Look at drongoski's solution =]

I don't get how he got sin x + cos x in the second line though.

 

[AnandDNA]

he had (1+cosx/sinx) he multiplied the 1 by sinx to have common denominator sinx

so it became (sinx+cosx)/sinx and this cancelled out with the 1/sinx

leaving you with sinx+cosx

lol sorry if my explanation was not understandable i couldnt be bothered working with Latex

 

[gurmies]

I don't get how he got sin x + cos x in the second line though.

Do the working out for the fraction and you'll notice that u have (sinx+cosx)/sinx/1/sinx = sinx + cosx.

 

[s2Vicki]

Thanks so much everyone. I get it now.