If $$\theta$$​ is an acute angle, and $$cos\theta = \frac{12}{13}$$​, then the value of $$\frac{13sin\theta + 12sec\theta}{12tan\theta + 5cosec\theta}$$​ is:

Correct option is A

Given:

​$$\cos \theta = \frac{12}{13}$$​​

We need to find the value of $$\frac{13 \sin \theta + 12 \sec \theta}{12 \tan \theta + 5 \csc \theta}$$​​

Concept Used:

Trigonometric identities:

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

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

​$$\cosec \theta = \frac{1}{\sin \theta}$$​​

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

Solution:

Using $$\cos \theta = \frac{12}{13}​,$$​​

​$$\sin^2 \theta = 1 - \cos^2 \theta = 1 - \left(\frac{12}{13}\right)^2 = 1 - \frac{144}{169} = \frac{25}{169}$$​​

Now, calculate the required trigonometric values:

​$$\sec \theta = \frac{1}{\cos \theta} = \frac{13}{12}$$​​

​$$\tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{5}{12} ​$$​​

​$$\cosec \theta = \frac{1}{\sin \theta} = \frac{13}{5} ​$$​​

Substituting values into the expression:

​$$\frac{13 \sin \theta + 12 \sec \theta}{12 \tan \theta + 5 \csc \theta} \\ \ \\ = \frac{13 \times \frac{5}{13} + 12 \times \frac{13}{12}}{12 \times \frac{5}{12} + 5 \times \frac{13}{5}} \\ \ \\= \frac{5 + 13}{5 + 13} = \frac{18}{18} = 1$$​​