If tan θ - cot θ = x and cos θ – sin θ = y, find the value of $$(x^2+4)(y^2-1)^2$$​.

Correct option is C

Given:
​$$\tan \theta - \cot \theta = x$$​​
​$$\cos \theta - \sin \theta = y$$

Formula Used:

$$sin^2\theta +cos^2\theta =1$$​​

1 + cot$$^2 \theta =cosec^2 \theta$$

tan $$\theta =\frac {sin\theta }{cos\theta}$$ ; cot $$\theta = \frac{cos \theta }{sin\theta}$$​​
Solution:
​$$x = \tan \theta - \cot \theta = \frac{\sin \theta}{\cos \theta} - \frac{\cos \theta}{\sin \theta} = \frac{\sin^2 \theta - \cos^2 \theta}{\sin \theta \cos \theta}$$​
​$$x = \frac{-2 \cos 2\theta}{2 \sin \theta \cos \theta} = \frac{-2 \cos 2\theta}{\sin 2\theta} = -2 \cot 2\theta$$​​

​$$x^2 + 4 = 4 \cot^2 2\theta + 4 = 4 (\cot^2 2\theta + 1) = 4 \cosec^2 2\theta$$ =$$\left(\frac{4}{\sin^2 2\theta}\right)$$​​
​$$y^2 -1= (\cos \theta - \sin \theta)^2 -1= \cos^2 \theta + \sin^2 \theta - 2 \sin \theta \cos \theta-1 = 1 - \sin 2\theta-1=-\sin 2\theta$$​​
​$$(x^2 + 4)(y^2 - 1)^2 = \left(\frac{4}{\sin^2 2\theta}\right) \times \sin^2 2\theta = 4$$​
Alternate Method:
Let $$\theta = \frac{\pi}{4}$$​​

x = $$\tan \frac{\pi}{4} - \cot \frac{\pi}{4} = 1 - 1 = 0$$​​
y =$$\cos \frac{\pi}{4} - \sin \frac{\pi}{4} = \frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}}= 0$$​
​$$(x^2 + 4)(y^2 - 1)^2 = (0 + 4)(0 - 1)^2 = 4 \times 1 = 4$$​