Verify cos(a+b)/cos(a) cos(b) =1-tan(a) tan(b)

One  Way:

Cos( a+b ) / cos(a) cos(b)

(cosacosb - sina sinb) / cosa cosb (as from formula of cos(a + b ) = cosacosb - sinasinb

we can spilt the above line as,

cosacosb /cosacosb - sinasinb / cosacosb ( dividing cosacosb by cosacosb we got 1)

= 1 - ( sina/cosa- sinb/cosb)( you know sina/cosa = tana ,   and sinb/cosb = tanb )

= 1 - tan(a) tan(b) = RHS ( right hand side)

Another Way:

(cos(a+b)/ cos(a)cos(b))= 1 -tan(a)tan(b)

”Sum of Cosine Identity”

cos(A + B) = cos A cos B − sin A sin B

(cos(a)cos(b) - sin (a) sin(b))/cos(a)cos(b)=...

(cos(a)cos(b))/(cos(a)cos(b)) - (sin(a)sin(b))/(cos(a)cos(b))=

sin(x)/cos(x) = tan(x) such that...

(cos(a)cos(b))/(cos(a)cos(b)) = 1

(sin(a)sin(b))/(cos(a)cos(b)) = tan(a)tan(b)

1 - tan(a)tan(b) = 1 - tan(a)tan(b)

ok so the answer to your question is simple as long as you look at the different dimensions that contain the exemplary letters that combined in the right order give you words that eventually within the 5th dimension will conform and give you the correct answer to your question. hope i , stop looking at this easily, look at it with a spectral view.

step-by-step explanation:

told you

x=0

step-by-step explanation: