Question

Prove the identity
((Cos²x - sin²x) ÷ (cosx + sinx) ) + sinx

Answer 1

Starting with the left side of the identity:

((cos²x - sin²x) ÷ (cosx + sinx)) + sinx

Using the identity cos²x = 1 - sin²x, we can substitute:

((1 - sin²x - sin²x) ÷ (cosx + sinx)) + sinx

Simplifying the numerator:

(1 - 2sin²x) ÷ (cosx + sinx) + sinx

Factoring out a negative 1 from the numerator:

- (2sin²x - 1) ÷ (cosx + sinx) + sinx

Using the identity sin²x + cos²x = 1, we can substitute:

- (2(1 - cos²x) - 1) ÷ (cosx + sinx) + sinx

Simplifying:

- (2 - 2cos²x - 1) ÷ (cosx + sinx) + sinx

Combining like terms:

- (1 - 2cos²x) ÷ (cosx + sinx) + sinx

Multiplying the numerator and denominator by -1 to flip the sign:

(2cos²x - 1) ÷ - (cosx + sinx) - sinx

Distributing the negative sign:

(2cos²x - 1) ÷ (-cosx - sinx) - sinx

Factoring out a -1 from the denominator:

(2cos²x - 1) ÷ -1(cosx + sinx) - sinx

Simplifying:

-(2cos²x - 1) ÷ (cosx + sinx) - sinx

Now, we can see that the left side of the identity simplifies to:

-(2cos²x - 1) ÷ (cosx + sinx) - sinx

which matches the expression on the right side of the identity. Therefore, the identity is proven.