Question

If `tan θ = p/q`, then prove that `(p sin θ - q cos θ)/(p sin θ + q cos θ) = ((p^2 - q^2)/(p^2 + q^2))`.

Answer 1

Given: `tan θ = p/q` with p and q not both zero.

To Prove: `(p sin θ - q cos θ)/(p sin θ + q cos θ) = (p^2 - q^2)/(p^2 + q^2)`

Proof [Step-wise]:

1. From `tan θ = p/q`, choose a right triangle or scale with opposite = p, adjacent = q.

So, the hypotenuse `r = sqrt(p^2 + q^2)`. 

This is valid since p and q are not both zero.

2. Then `sin θ = p/r` and `cos θ = q/r`.

3. Compute the numerator:

p sin θ – q cos θ 

= `p xx (p/r) - q xx (q/r)` 

= `(p^2 - q^2)/r`

4. Compute the denominator:

p sin θ + q cos θ

= `p xx (p/r) + q xx (q/r)`

= `(p^2 + q^2)/r`

5. Form the ratio:

`(p sin θ - q cos θ)/(p sin θ + q cos θ)`

= `[(p^2  -  q^2)/r]/[(p^2  +  q^2)/r]`

= `(p^2 - q^2)/(p^2 + q^2)`

Hence, given `tan θ = p/q, (p sin θ - q cos θ)/(p sin θ + q cos θ) = ((p^2 - q^2)/(p^2 + q^2))`.