The value of 
​$$\cot18^o\left[\cot72^o . \cos^222^o + \frac{1}{\tan72^o . \sec^268^o}\right]$$​​

Correct option is C

Given: ​

​$$\cot18^o\left[\cot72^o . \cos^222^o + \frac{1}{\tan72^o . \sec^268^o}\right]$$ 

Formula Used:  

​$$\cot(90^0 - \theta) = \tan \theta \\ \ \\ \sin^2 \theta +\cos^2\theta = 1$$ 

Solution: ​

​$$\cot18^o\left[\cot72^o . \cos^222^o + \frac{1}{\tan72^o . \sec^268^o}\right] \\ \ \\ = \cot18^o\left[\cot72^o . \cos^222^o + \cot72^o . \cos^268^o\right] \\ \ \\ = \cot18^o\left[\cot72^o . \cos^2(90^o -68^o) + \cot72^o . \cos^268^o\right] \\ \ \\ =\cot18^o\left[\cot72^o . \sin^268^o + \cot72^o . \cos^268^o\right] \\ \ \\ = \cot18^o\left[\cot72^o ( \sin^268^o + \cos^268^o)\right]\\ \ \\ = \cot18^o\left[\cot72^o ( 1)\right] \\ \ \\ = \tan 72^0 \times \frac{1}{\tan 72^0}=1$$​​