If $$x cosθ - sinθ = 1,$$ then $$x^{2} - (1 +x^{2}) sinθ$$ equals

Expression : $$x cosθ - sinθ = 1$$

=> $$x = \frac{1}{cos\theta} + \frac{sin\theta}{cos\theta}$$

=> $$x = sec\theta + tan\theta$$ --------------Eqn(1)

$$\because$$ $$sec^2\theta - tan^2\theta = 1$$

=> $$(sec\theta + tan\theta)(sec\theta - tan\theta) = 1$$

=> $$(sec\theta - tan\theta) = \frac{1}{x}$$ --------------Eqn(2)

Adding eqns(1)&(2)

=> $$2sec\theta = x + \frac{1}{x} = \frac{x^2 + 1}{x}$$

=> $$sec\theta = \frac{x^2 + 1}{2x}$$

Subtracting eqn(2) from (1)

=> $$2tan\theta = x - \frac{1}{x} = \frac{x^2 - 1}{x}$$

=> $$tan\theta = \frac{x^2 - 1}{2x}$$

We know that, $$sin\theta = \frac{tan\theta}{sec\theta}$$

=> $$sin\theta = \frac{x^2 - 1}{2x} * \frac{2x}{x^2 + 1}$$

=> $$sin\theta = \frac{x^2 - 1}{x^2 + 1}$$

To find : $$x^2 - (1 + x^2) sinθ $$

= $$x^2 - (1 + x^2) * \frac{x^2 - 1}{x^2 + 1}$$

= $$x^2 - (x^2 - 1)$$

= 1