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

Answer 1

Hi Bella, as I understand from the above sum You have to prove the right hand side is equal to the Left Hand side.

Lets starts from LHS ( left hand side),

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) Proved Ans

Answer 2

Proof:

(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)

it works