Question

if cos 42 degrees=a then find the value of tan 48 degrees

Answer 1

Answer on Question #57361 – Math – Trigonometry

Question

If $\cos 42$ degrees = a then find the value of $\tan 48$ degrees.

Solution

It is known that

$$\cot \alpha = \tan (90{}^\circ - \alpha)$$ $$\tan \alpha = \cot (90{}^\circ - \alpha)$$

So

$$\tan 48{}^\circ = \cot (90{}^\circ - 48{}^\circ)$$ $$\tan 48{}^\circ = \cot 42{}^\circ.$$

The another identities show that

$$\cot \alpha = \frac{\cos \alpha}{\sin \alpha}$$

and

$$\sin^2 \alpha + \cos^2 \alpha = 1,$$

hence

$$\sin \alpha = \sqrt{1 - \cos^2 \alpha}, \text{ when } 0{}^\circ < \alpha < 90{}^\circ.$$

From this we can derive:

$$\cot \alpha = \frac{\cos \alpha}{\sqrt{1 - \cos^2 \alpha}}.$$

So, we substitute $\alpha = 42{}^\circ$ and get the formula:

$$\cot 42{}^\circ = \frac{\cos 42{}^\circ}{\sqrt{1 - \cos^2 42{}^\circ}}$$ $$\cot 42{}^\circ = \frac{a}{\sqrt{1 - a^2}}.$$

Because

$$\tan 48{}^\circ = \cot 42{}^\circ,$$

the answer will be

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

Answer: $\tan 48{}^\circ = \frac{a}{\sqrt{1 - a^2}}$.

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