Sinα +sinβ cos(α +β) /cosα −sinβ sin(α +β) b) = tan(α +β)

To prove that sinα + sinβ cos(α + β) / cosα − sinβ sin(α + β) = tan(α + β), we can start by simplifying the left side of the equation.

Using the trigonometric identity sin(α + β) = sinα cosβ + cosα sinβ, we can rewrite the numerator as:

sinα + sinβ cos(α + β) = sinα + sinβ (cosα cosβ - sinα sinβ)

Expanding the expression, we get:

sinα + sinβ cos(α + β) = sinα + sinβ cosα cosβ - sinβ sinα sinβ

Now, let's simplify the denominator using the trigonometric identity cos(α + β) = cosα cosβ - sinα sinβ:

cosα − sinβ sin(α + β) = cosα − sinβ (sinα cosβ + cosα sinβ)

Expanding the expression, we get:

cosα − sinβ sin(α + β) = cosα − sinβ sinα cosβ - sinβ cosα sinβ

Now, let's substitute these simplified expressions back into the original equation:

(sinα + sinβ cosα cosβ - sinβ sinα sinβ) / (cosα − sinβ sinα cosβ - sinβ cosα sinβ)

To simplify this expression further, we can factor out sinβ from the numerator and denominator:

sinβ (sinα cosα cosβ - sinα sinβ - cosα sinβ) / sinβ (cosα - sinα cosβ - cosα sinβ)

Now, we can cancel out sinβ from the numerator and denominator:

(sinα cosα cosβ - sinα sinβ - cosα sinβ) / (cosα - sinα cosβ - cosα sinβ)

Now, let's simplify the numerator:

sinα cosα cosβ - sinα sinβ - cosα sinβ = sinα (cosα cosβ - sinβ) - sinβ (cosα + sinα)

Now, let's simplify the denominator:

cosα - sinα cosβ - cosα sinβ = cosα (1 - sinβ) - sinα cosβ

Now, let's substitute these simplified expressions back into the equation:

(sinα (cosα cosβ - sinβ) - sinβ (cosα + sinα)) / (cosα (1 - sinβ) - sinα cosβ)

Expanding the expression, we get:

(sinα cosα cosβ - sinα sinβ - sinβ cosα - sinβ sinα) / (cosα - cosα sinβ - sinα cosβ)

Now, let's simplify the numerator further:

sinα cosα cosβ - sinα sinβ - sinβ cosα - sinβ sinα = 0

Therefore, the left side of the equation simplifies to 0.

Since tan(α + β) is not equal to 0, we can conclude that sinα + sinβ cos(α + β) / cosα − sinβ sin(α + β) is not equal to tan(α + β).

Hence, the given equation is not true.