Question

Prove that `1 - (sin(90^circ - θ) sin θ)/(tan θ) = sin^2θ`.

Answer 1

Given: `1 - (sin (90^circ - θ) · sin θ)/(tan θ)`

To Prove: `1 - (sin (90^circ - θ) · sin θ)/(tan θ) = sin^2θ`.

Proof [Step-wise]:

1. Use the co-function identity:

sin (90° – θ) = cos θ

Substitute into the expression:

`1 - (cos θ · sin θ)/(tan θ)`

2. Replace tan θ by `(sin θ)/(cos θ)`: 

`(cos θ · sin θ)/(tan θ) = (cos θ · sin θ)/((sin θ)/(cos θ))`

3. Simplify the fraction cancel sin θ; allowed because the expression requires tan θ ≠ 0, so sin θ ≠ 0: 

`(cos θ · sin θ) · ((cos θ)/(sin θ)) = cos^2θ`

4. Therefore, the whole expression becomes 1 – cos²θ.

5. Apply the Pythagorean identity 1 – cos²θ = sin²θ.

Hence, `1 - (sin(90^circ - θ) · sin θ)/tan θ = sin^2θ`.

`1 - (sin(90^circ - θ) sin θ)/tan θ = sin^2θ` for θ with tan θ defined and ≠ 0.