prove:
sin(a+b)/cosacosb=tana+tanb
sin(a+b)/cos(a)cos(b) =
= (sin(a)cos(b)+cos(a)sin(b))/cos(a)cos(b)
= sin(a)cos(b)/(cos(a)cos(b) +
cos(a)sin(b)/(cos(a)cos(b)
= sin(a)/cos(a) + sin(b)/cos(b)
= tan(a) + tan(b)
To prove the equality sin(a+b) / (cos(a)cos(b)) = tan(a) + tan(b), we can start by using the trigonometric identities for sine, cosine, and tangent:
sin(a + b) = sin(a)cos(b) + cos(a)sin(b)
cos(a + b) = cos(a)cos(b) - sin(a)sin(b)
Then, we can substitute these identities into the given equation:
sin(a + b) / (cos(a)cos(b)) = (sin(a)cos(b) + cos(a)sin(b)) / (cos(a)cos(b))
Next, we can simplify the numerator by factoring out sin(b) in the numerator:
= (sin(b)(cos(a) + sin(a))) / (cos(a)cos(b))
Now, we can cancel out the common factors of cos(a) in the numerator and denominator:
= sin(b) / cos(b)
Finally, we can use the identity of tan(x) = sin(x) / cos(x) to rewrite the expression:
= tan(b)
Therefore, sin(a+b) / (cos(a)cos(b)) = tan(a) + tan(b).
To prove the equation:
sin(a+b) / (cosa * cosb) = tana + tanb
We can start by using the formula for the sine of the sum of angles:
sin(a+b) = sina*cosb + cosa*sinb
Substituting this back into the equation, we get:
(sina*cosb + cosa*sinb) / (cosa * cosb) = tana + tanb
Next, we simplify the expression by canceling out common factors:
(sina*cosb) / (cosa * cosb) + (cosa*sinb) / (cosa * cosb) = tana + tanb
Now, we can simplify further by canceling out the common factors:
sin(a)/cos(a) + sin(b)/cos(b) = tana + tanb
Recall that sin(a)/cos(a) is equal to tan(a), and sin(b)/cos(b) is equal to tan(b). Rewriting the equation:
tana + tanb = tana + tanb
Therefore, we have shown that sin(a+b) / (cosa * cosb) is indeed equal to tana + tanb.