If 16 $$\text{sec}^2$$​θ – 40 secθ + 25 = 0 and θ is an acute angle, then what will be the value of tan θ?

Correct option is A

Given:

​$$16 \sec^2 \theta - 40 \sec \theta$$​ + 25 = 0

Concept Used:

​$$\sec^2 \theta = 1 + \tan^2 \theta:$$​​

Solution:

​$$16 \sec^2 \theta - 40 \sec \theta + 25$$​= 0

16$$\sec^2\theta -20\sec\theta -20\sec\theta +25$$​ = 0

4$$\sec\theta(4\sec\theta-5)-5 (4\sec\theta-5)=$$​0

​$$(4\sec\theta-5) (4\sec\theta-5)$$​

4$$\sec\theta = 5$$​​

​$$\sec\theta = \frac54$$​

$$\sec^2 \theta = 1 + \tan^2 \theta:$$​

​$$\left(\frac{5}{4}\right)^2 = 1 + \tan^2 \theta$$​

$$\frac{25}{16} = 1 + \tan^2 \theta\\ \ \\\frac{25}{16} - 1 = \tan^2 \theta\\ \ \\\frac{25}{16} - \frac{16}{16} = \tan^2 \theta\\ \ \\\frac{9}{16} = \tan^2 \theta\\ \ \\\tan \theta = \frac{3}{4}$$​