Stumped on the verification of a purely trigonometric identity

[Al-Layth]

the trigonometric identity to be proved is:

[math]\frac{2+2 \sin(x)}{\cos(x) + \sin(x) \cos(x)} = \frac{2}{\cos(x)}[/math]
my general approach is to try and simplify the more complicated side, and recognise any expressions that fit within the body of trigonometric identities known to me

but I am completely stumped here. I can see that I already have the goal expression embedded in the LHS which is my working side, but the extra terms suggest the identity is false??

I recognise the sin(x)cos(x) as writable in terms of sin(2x) but introducing a double angle into a single angle equation cant help.

thoughts?

 

[blamocur]

If [imath]\frac{a}{b} = \frac{c}{d}[/imath] then [imath]ad = bc[/imath].

 

[Al-Layth]

If [imath]\frac{a}{b} = \frac{c}{d}[/imath] then [imath]ad = bc[/imath].

how do i apply this esoteric fact here?

 

[BigBeachBanana]

the trigonometric identity to be proved is:

[math]\frac{2+2 \sin(x)}{\cos(x) + \sin(x) \cos(x)} = \frac{2}{\cos(x)}[/math]
my general approach is to try and simplify the more complicated side, and recognise any expressions that fit within the body of trigonometric identities known to me

but I am completely stumped here. I can see that I already have the goal expression embedded in the LHS which is my working side, but the extra terms suggest the identity is false??

I recognise the sin(x)cos(x) as writable in terms of sin(2x) but introducing a double angle into a single angle equation cant help.

thoughts?

Factor out the GCF of the numerator and the denominator of the LHS.

 

[Steven G]

As just mentioned, factor the numerator and denominator and you'll be done.

 

[Al-Layth]

Factor out the GCF of the numerator and the denominator of the LHS.

Damn how did i not see that

thank you

 

[Harry_the_cat]

Damn how did i not see that

thank you

Always look for the simplest way!