If x = (sec⁡θ-cos⁡θ)(cot⁡θ+tan⁡θ) and y = sec⁡θ(sec⁡θ+tan⁡θ)(1-sin⁡θ), $$0^∘<θ<90^∘$$​, then xy is equal to:

Correct option is A

Given:

x = (secθ-cosθ)(cotθ+tanθ)

y = secθ(secθ+tanθ)(1-sinθ)

Formula Used:

​$$\sec \theta = \frac{1}{\cos\theta}$$​​

​$$\tan\theta = \frac{\sin\theta}{\cos\theta}$$​​

​$$\cot\theta = \frac{\cos\theta}{\sin\theta}$$​​

​$$\sin^2 \theta + \cos^2 \theta = 1$$​​

Solution:

​$$xy=(secθ-cosθ)(cotθ+tanθ) \times secθ(secθ+tanθ)(1-sinθ)$$​​

​$$= \left(\frac{1}{cos \theta} - \cos \theta\right)\left(\frac{\cos \theta}{\sin \theta} + \frac{\sin \theta}{\cos \theta}\right) \times \frac{1}{\cos \theta} \left(\frac{1}{\cos \theta} + \frac{\sin \theta}{\cos \theta}\right) (1- \sin \theta)$$​​

​$$= \left(\frac{1 - \cos^2 \theta}{cos \theta}\right)\left(\frac{\cos^2 \theta + \sin^2 \theta}{\sin \theta \cos \theta}\right) \times \frac{1}{\cos \theta} \left(\frac{1+ \sin \theta}{\cos \theta} \right) (1- \sin \theta)$$​​

​$$= \left(\frac{\sin^2 \theta}{cos \theta}\right)\left(\frac{1}{\sin \theta \cos \theta}\right) \times \left(\frac{1+ \sin \theta}{\cos^2 \theta} \right) (1- \sin \theta)$$​​

​$$= \left(\frac{\sin \theta}{cos^2 \theta}\right) \times \left(\frac{1+ \sin \theta}{1 - \sin^2 \theta} \right) (1- \sin \theta)$$​​

​$$= \left(\frac{\sin \theta}{cos^2 \theta}\right) \times \frac{1+ \sin \theta}{(1+ \sin \theta)(1- \sin \theta)} (1- \sin \theta)$$​​

​$$= \tan \theta \frac{1}{\cos \theta} \times 1$$​​

​$$= \tan \theta \sec \theta$$​​