Question

1/sinx + 1/cosx=(cosx+sinx)(secx)(cscx) @zzr0ck3r

Answer 1

@zzr0ck3r

Answer 2

distribute the secx and cscx on the right and tell me what you have

Answer 3

Cupcake: pretty please, include instructions when you post a problem.
Have you tried adding the two fractions on the left AFTER finding and using the LCD?

Answer 4

??@mathmale

Answer 5

I need help with this...

Answer 6

@zzr0ck3r Idk what I even did please help

Answer 7

@zz0ck3r help

Answer 8

you have (a+b)*c*d this is the same thing as acd+bcd
I distributed the cd
Do the same thing with sec(x)csc(x)

Answer 9

what do you get?

Answer 10

Starting with 1/sinx + 1/cosx=(cosx+sinx)(secx)(cscx)
I'd write this a bit differently for simplicity:
\[\frac{ 1 }{ \sin x }+\frac{ 1 }{ \cos x }\]

Answer 11

Then I'd identify the LCD (lowest common denominator) of these two fractions. What is it?
Combine these two fractions using the LCD. Please show your work so i can give you feedback on it.

Answer 12

Come on!! To find the LCD, just multiply: (sin x)(cos x).

Then re-write that sum of fractions as \[\frac{ \cos x }{ (\sin x)(\cos x) }+\frac{ \sin x }{ (\sin x)(\cos x) }.\]

what next in the problem solution? (Hint: try looking up trig identities and applying the appropriate ones.)