In a right angle triangle ABC, $$\angle C = 90\degree$$, if sinA cosB = $$\frac{1}{4}$$​ and 3tanA = tanB, then $$\text{cot}^2A$$​ is equal to:

Correct option is A

Given:

​$$\sin A \cdot \cos B = \frac{1}{4}, \quad \text{and} \quad 3\tan A = \tan B$$​​

Concept Used:

​$$\cot^2 A = \frac{1}{\tan^2 A}$$​​

​$$\sin A = \frac{\tan A}{\sqrt{1 + \tan^2 A}},$$​​

​$$\cos B = \frac{1}{\sqrt{1 + \tan^2 B}}$$​​

Solution:

Alternate Method:

Since, A + B = 90$$\degree$$ as ABC is a triangle

A = 90$$\degree$$ - B

sin A = sin (90$$\degree$$​ - B)

sin A =  cos B

So, $$\sin A \cdot \cos B = \frac{1}{4}$$

sin² A = $$\frac14$$

sin A = $$\frac 12$$

A = 30$$\degree$$

Then, cot² A = cot ²30$$\degree$$= $$(\sqrt 3)^2=3$$​​