If cos 42° = p, then tan 48° = ?

Correct option is B

Given:

​$$\cos 42^\circ = p$$​​

Concept Used:

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

$$\cot \theta = \frac{1}{\tan \theta}$$

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

Solution:

​$$\tan 48^\circ = \cot 42^\circ$$​​

Hence,

​$$\tan 48^\circ = \frac{\cos 42^\circ}{\sin 42^\circ}$$​​

Now, since $$\cos 42^\circ = p$$​, we can use the identity:

​$$\sin^2 42^\circ = 1 - \cos^2 42^\circ$$​​

Thus,

​$$\sin 42^\circ = \sqrt{1 - p^2}$$​​

Therefore,

$$\tan 48^\circ = \frac{p}{\sqrt{1 - p^2}}$$

Alternate Method:

​$$\cos 42^\circ = p$$

​​$$\sin 48^\circ =\frac p1 =\frac {\text{Perpendicular}}{\text{Hypotenuse}}$$ 

By Pythagorean Theorem

Perpendicular² + Base² = Hypotenuse²​

Base = $$\sqrt{1-p^2}$$   

​tan$$48^\circ$$ = $$\frac {\text{Perpendicular}}{\text{Base}}$$ = $$\frac{p}{\sqrt{1-p^2}}$$​​