prove tanA+tanB=sin(A+B)/cosAcosB
sin(A+B) = sinAcosB+cosAsinB
divide that by cosAcosB and you have
sinA/cosA + sinB/cosB
elaborate more
To prove the identity tan A + tan B = sin(A + B) / (cos A * cos B), we can start with the definition of tangent.
The tangent of an angle A is defined as the ratio of the sine of that angle to the cosine of that angle:
tan A = sin A / cos A
Similarly, the tangent of an angle B is defined as the ratio of the sine of angle B to the cosine of angle B:
tan B = sin B / cos B
Now, we can rewrite the left-hand side (LHS) of the equation:
tan A + tan B = (sin A / cos A) + (sin B / cos B)
To combine these two fractions, we need to first find a common denominator. The denominator can be found by multiplying the denominators of the two fractions together, which gives us:
cos A * cos B
Now, we can rewrite the fractions with the common denominator:
tan A + tan B = (sin A * cos B) / (cos A * cos B) + (sin B * cos A) / (cos B * cos A)
Notice that the numerators of the two fractions have been multiplied by the denominator that was initially in the other fraction. This is done to ensure that we maintain the same value while also finding the common denominator.
Simplifying the expression, we have:
tan A + tan B = (sin A * cos B + sin B * cos A) / (cos A * cos B)
Now, let's focus on the numerator of this expression. Using the sum formula for sines, we can rewrite the numerator as:
sin A * cos B + sin B * cos A = sin(A + B)
Hence, we have:
tan A + tan B = sin(A + B) / (cos A * cos B)
And this proves the given identity.